457. Relatii metrice si inegalitati geometrice.

by Virgil Nicula, Aug 19, 2017, 11:10 AM

P5 (Miguel Ochoa Sanchez). Prove that for any $\triangle ABC$ there is the chain of the inequalities $:\ \boxed{4\left(R^2-r^2\right)\le HA^2+HB^2+HC^2\le 12(R-r)^2}$ (standard notations).

Proof 1 (metric). Denote the midpoint $M$ of the side $[BC]$ and the diameter $[AS]$ of the circumcircle $w=\mathbb C(O,R).$ Apply the theorem of median in the triangles

$\left\{\begin{array}{cccc} \triangle BHC\ : & 2\left(HB^2+HC^2\right) & = & HS^2+BC^2\\\\ \triangle HAC\ : & 4AM^2+HS^2 & = & 2\left(HA^2+AS^2\right)\end{array}\right\|\bigoplus\implies$ $2\left(HB^2+HC^2-HA^2\right)+4m_a^2-a^2=$ $8R^2\iff$ $HB^2+HC^2-HA^2+$ $b^2+c^2-a^2=$ $4R^2$

$\iff$ $\sum HA^2+\sum a^2=12R^2\iff$ $\boxed{\sum HA^2=12R^2-2\left(s^2-r^2-4Rr\right)}\ (1).$ I"ll prove that the required bilateral inequality is equivalently with the Gerretsen's inequality:

$\boxed{16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2}\ (*).$ Indeed: $4\left(R^2-r^2\right)\le \sum HA^2\le 12(R-r)^2\ \stackrel{1}{\iff}\ 4\left(R^2-r^2\right)\le 12R^2-2\left(s^2-r^2-4Rr\right)\le 12(R-r)^2\iff$

$2\left(R^2-r^2\right)\le 6R^2-s^2+r^2+4Rr\le 6(R-r)^2\iff$ $6R^2+r^2+4Rr-6(R-r)^2\le s^2\le 6R^2+r^2+4Rr-2\left(R^2-r^2\right)\iff$ the bilateral inequality $(*)\ .$

Proof 2 (trigonometric). I"ll use the wellknown identity $HA=2R|\cos A|$ a.s.o. Thus $\sum HA^2=4R^2\cdot \sum\cos ^2A=$ $4R^2\cdot\sum\left(1-\sin^2A\right)=$ $12R^2-\sum a^2=$

$12R^2-2\left(s^2-r^2-4Rr\right)\implies$ $\sum HA^2=12R^2-2\left(s^2-r^2-4Rr\right)$ i.e. the relation $(1)$ a.s.o.

Proof 3 (metric). Denote the diameter $[AN]$ of the circumcircle $w=\mathbb C(O,R).$ Prove easily that $AHCN$ is a parallelogram, i.e. $HA=NC$ and $NC^2+BC^2=BN^2\implies$

$\boxed{HA^2+a^2=4R^2}$ a.s.o. In conclusion, $\boxed{\sum HA^2+\sum a^2=12R^2}\ (2)\implies \sum HA^2=12R^2-2\left(s^2-r^2-4Rr\right)$ i.e. the relation $(1)$ a.s.o.

Proof 4 (metric). Apply the remarkable identity $\sum XA^2=3\cdot XG^2-3\mathrm p_{w}(G)$ for the centroid $H\ :\ \sum HA^2=3\cdot HG^2+\frac 13\cdot\sum a^2=$ $3\cdot \left(2\cdot OG\right)^2+\frac 13\cdot \sum a^2=$

$12\cdot OG^2+\frac 13\cdot\sum a^2=$ $12\cdot \left(R^2-\frac 19\cdot\sum a^2\right)+\frac 13\cdot \sum a^2=$ $12R^2-\sum a^2\implies$ $\sum HA^2+\sum a^2=12R^2,$ i.e. the relation $(2)$ a.s.o.

P6. Prove that for any $\triangle ABC$ there is the Petrovic's inequality $:\ \boxed{\frac{\sqrt 3}R\le \frac 1a+\frac 1b+\frac 1c\le \frac {\sqrt 3}{2r}}$ , i.e. $\boxed{\frac 1{h_a}+\frac 1{h_b}+\frac 1{h_c}\ge \frac 2{\sqrt 3}\cdot \left(\frac 1a+\frac 1b+\frac 1c\right)}\ (*)$ (standard notations).

Proof. Observe that $\sum\frac 1{h_a}=\sum\frac {a}{ah_a}=\frac sS=\frac 1r$ . Thus, the relation $(*)$ is true $\iff$ the Petrovic's relation is true $\iff$ $\sum\frac {h_a}{2S}\le \frac {\sqrt 3}{2r}\iff$ $\boxed{h_a+h_b+h_c\le s\sqrt 3}\ (1)$ $\iff$

$\boxed{\frac {ab+bc+ca}{a+b+c}\le R\sqrt 3}\ (2)$ . I used the well-known identity $bc=2Rh_a$ . Since $\frac {ab+bc+ca}{a+b+c}\le$ $\frac {a+b+c}3\le$ $\frac {2s}3\le R\sqrt 3$ get easily the relation $(2)$ , i.e. the required inequality $(*)$ .

Here is a proof of the (left) inequality $\frac {\sqrt 3}{R}\le\frac 1a+\frac 1b+\frac 1c$ . Indeed, with remarkable relations $\left\{\begin{array}{ccc} \frac sr & \ge & 3\sqrt 3\\\\ \frac {4R+r}s & \ge & \sqrt 3\end{array}\right\|$ get $\sum\frac 1a=\frac {ab+bc+ca}{abc}=\frac {s^2+r^2+4Rr}{4Rsr}=$

$\frac 1s+\frac 1{4R}\cdot\left(\frac sr+\frac rs\right)\ge$ $\frac 1s+\frac 1{4R}\left(3\sqrt 3+\frac rs\right)=$ $\frac {3\sqrt 3}{4R}+\frac 1s\left(1+\frac r{4R}\right)=$ $\frac {3\sqrt 3}{4R}+\frac {4R+r}s\cdot \frac 1{4R}\ge$ $\frac {3\sqrt 3}{4R}+\frac {\sqrt 3}{4R}=\frac {\sqrt 3}R$ . In conclusion, $\boxed{\frac {\sqrt 3}{R}\le\frac 1a+\frac 1b+\frac 1c\le\frac {\sqrt 3}{2r}}\ .$

P7. Prove that in any triangle $ABC$ there is the inequality $\boxed{\frac {a+b+c}{2R-r}\le \frac {a}{r_a}+\frac {b}{r_b}+\frac {c}{r_c}\le \frac {a+b+c}{3r}}$ .

Proof.

Proof. Is well-known that $ar_a+br_b+cr_c=2s(2R-r)$ , where $a+b+c=2s$ . Apply the CBS inequality $\sum\frac{a}{r_a}=\sum \frac {a^2}{ar_a}\ge$

$\frac{(a+b+c)^2}{ar_a+br_b+cr_c}=\frac{4s^2}{2s(2R-r)}=\frac {2s}{2R-r}\implies$ $\frac {a+b+c}{2R-r}\le \frac {a}{r_a}+\frac {b}{r_b}+\frac {c}{r_c}$ . Observe that $a\le b\le c\iff$ $\frac 1{r_a}\ge\frac 1{r_b}\ge\frac 1{r_c}$ .

Hence can apply the Chebyshev's inequality $:\ \sum\frac a{r_a}=\sum\left( a\cdot\frac 1{r_a}\right)\le \frac 13\cdot\sum a\cdot\sum\frac 1{r_a}=\frac {2s}{3r}\implies$ $\frac {a}{r_a}+\frac {b}{r_b}+\frac {c}{r_c}\le \frac {a+b+c}{3r}$ .

Remark. $\left(r_a+r_b+r_c\right)^2\ge 3\left(r_ar_b+r_br_c+r_cr_a\right)\iff$ $(4R+r)^2\ge 3s^2\iff$ $\boxed{s\sqrt 3\le 4R+r}\ (*)$ . Thus, $\boxed{9r\le s\sqrt 3\le 4R+r\le \frac {9R}2}$ .

P8. Prove that in any triangle $ABC$ there is the inequality $\boxed{\frac {a}{b+c-a}+\frac {b}{a-b+c}+\frac {c}{a+b-c}\ge 3}$ .

Proof 1. Denote $\left\{\begin{array}{c} s-a=x>0\\\\ s-b=y>0\\\\ s-c=z>0\end{array}\right\|$ , where $2s=a+b+c$ . Thus, our inequality becomes $\sum\frac {y+z}{x}\ge 6\iff$ $\left(\frac xy+\frac yx\right)+\left(\frac yz+\frac zy\right)+\left(\frac zx+\frac xz\right)\ge 6$ , what is truly.

Proof 2. $\sum\frac a{b+c-a}=$ $\sum\frac {a^2}{a(b+c-a)}\ \stackrel{(CBS)}{\ge}\ \frac {(a+b+c)^2}{\sum [a(b+c-a)]}=$ $\frac {(a+b+c)^2}{ab+bc+ca}\ge 3$ because $(a+b+c)^2\ge 3(ab+bc+ca)$ is well-known.

Remark. $\sum\frac a{b+c-a}=$ $\sum\frac {a^2}{a(b+c-a)}\ \stackrel{(CBS)}{\ge}\ \frac {(a+b+c)^2}{\sum [a(b+c-a)]}=$ $\frac {(a+b+c)^2}{ab+bc+ca}=$ $\frac {4s^2}{s^2+r(4R+r)}=$ $4-\frac {4r(4R+r)}{s^2+r(4R+r)}\ge$

$4-\frac {4r(4R+r)}{(16Rr-5r^2)+r(4R+r)}=$ $4-\frac {4r(4R+r)}{20Rr-4r^2}=$ $4-\frac {4R+r}{5R-r}=$ $\frac {16R-5r}{5R-r}=$ $3+\frac {R-2r}{5R-r}\ge 3$ because $R\ge 2r$ .

I used the well-known inequality $s^2+5r^2\ge 16Rr$ . In conclusion, we obtained the stronger inequality $\boxed{\sum\frac a{b+c-a}\ge \frac {16R-5r}{5R-r}}\ge 3$ .

P9. Prove that in any triangle $ABC$ there is the inequality $\boxed{4(a+b+c)\le\frac {(b+c)^2}{-a+b+c}+\frac {(c+a)^2}{a-b+c}+\frac {(a+b)^2}{a+b-c}\le \frac {4abc(a+b+c)}{(-a+b+c)(a-b+c)(a+b-c)}}$ .

Proof. $\sum\frac {(b+c)^2}{-a+b+c}\ \stackrel{(CBS)}{\ge}\ \frac {\left[\sum (b+c)\right]^2}{\sum (-a+b+c)}=$ $\frac {4(a+b+c)^2}{a+b+c}\implies$ $\boxed{\sum\frac {(b+c)^2}{-a+b+c}\ge 4(a+b+c)}\ (1)$ . Denote $\left\{\begin{array}{c} s-a=x>0\\\\ s-b=y>0\\\\ s-c=z>0\end{array}\right\|\implies$ $\left\{\begin{array}{c} a=y+z\\\\ b=z+x\\\\ c=x+y\end{array}\right\|$ .

Thus, $\sum\frac {(b+c)^2}{-a+b+c}=\sum\frac {(y+z+2x)^2}{2x}=$ $\sum\left[2x+2(y+z)+\frac {(y+z)^2}{2x}\right]\implies$ $\sum\frac {(b+c)^2}{-a+b+c}=6(x+y+z)+\frac w2$ , where $\boxed{w=\sum\frac {(y+z)^2}x}$ a.s.o.

Otherwise. $\sum {\frac{(a+b)^2}{a+b-c}}=$ $\sum {\frac{16R^2\cos^2\frac{C}{2}\cos^2\frac{A-B}{2}}{8R\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}}}\le$ $\sum \frac{2R\cos\frac{C}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}=$ $\frac{2R\sum {\sin\frac{A}{2}\cos\frac{A}{2}}}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=$ $\frac{R(\sin A+\sin B+\sin C)}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=$ $\frac{2R(a+b+c)}{r}=$ $\frac {4Rs}{r}=$

$\frac {4Rsr}{r^2}=\frac {abc}{r^2}=$ $\frac {4abc(a+b+c)}{8sr^2}\implies$ $\boxed{\sum {\frac{(b+c)^2}{-a+b+c}}\le\frac {4abc(a+b+c)}{(-a+b+c)(a-b+c)(a+b-c)}}\ (2)$ because $\left\{\begin{array}{ccc} \cos^2\frac{A-B}2 & \le & 1\\\\ \sin A+\sin B+\sin C & = & \frac sR\\\\ \sin\frac A2\sin\frac B2\sin\frac C2 & = & \frac r{4R}\end{array}\right\|$ .

P10. Prove that $\{x,y,z\}\subset\mathbb R^*_+\implies$ $7xyz+3\sqrt {\left(x^2+y^2+z^2\right)\left(x^4+y^4+z^4\right)}\ge 2(x+y)(y+z)(z+x)$ .

Proof. I"ll use the well-known identity $(x+y+z)^3=(x^3+y^3+z^3)+3(x+y)(y+z)(z+x)$ . Our inequality is equivalently with

$21xyz+9\sqrt{(\sum_{cyc}x^2)\cdot(\sum_{cyc}x^4)}\geq 2(x+y+z)^3-2(x^3+y^3+z^3)$ . From the Cauchy-Buniakowski-Schwarz's inequality obtain that $21xyz+9\sqrt{(\sum_{cyc}x^2)(\sum_{cyc}x^4)}\geq$

$21xyz+9(x^3+y^3+z^3)\geq$ $2(x+y+z)^3-2(x^3+y^3+z^3)\Leftrightarrow$ $21xyz+11(x^3+y^3+z^3)\geq$ $2(x+y+z)^3=$ $2(x^3+y^3+z^3)+6\sum_{cyc}ab(a+b)+12xyz\Leftrightarrow$

$9xyz+9(x^3+y^3+z^3)\geq$ $6\sum_{cyc}ab(a+b)\Leftrightarrow$ $(3xyz+x^3+y^3+z^3)+ 2(x^3+y^3+z^3)\geq$ $2\sum_{cyc}ab(a+b)$ . From the Schur's inequality for $r=1$ obtain that

$3xyz+x^3+y^3+z^3\geq\sum_{cyc}ab(a+b)\Leftrightarrow \sum_{cyc}a(a-b)(a-c)\geq 0$ . Remain to prove $2(x^3+y^3+z^3)\ge$ $\sum_{cyc}ab(a+b)$ what results from $a^3+b^3\geq ab(a+b), (\forall) a,b\in\mathbb{R^*_{+}}$

P11. Prove that $\{a,b,c\}\subset\mathbb R_+^*\ \Longrightarrow\ \left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge abc(a+b)(b+c)(c+a)$ and $\left(a^2+c^2\right)\left(b^2+c^2\right)\geq c(a+b)(ab+c^2)$ .

Proof 1. . $\left\|\ \begin{array}{c} a^4+b^2c^2\ge 2a^2bc\\\\ a^2\left(b^2+c^2\right)=a^2\left(b^2+c^2\right)\end{array}\ \right\|\ \bigoplus\ \Longrightarrow\ \left(a^2+b^2\right)\left(a^2+c^2\right)\ge$ $a^2(b+c)^2 \implies$ $\left\|\ \begin{array}{c} \left(a^2+b^2\right)\left(a^2+c^2\right)\ge a^2(b+c)^2\\\\ \left(b^2+c^2\right)\left(b^2+a^2\right)\ge b^2(c+a)^2\\\\ \left(c^2+a^2\right)\left(c^2+b^2\right)\ge c^2(a+b)^2\end{array}\ \right\|\ \bigodot\ \Longrightarrow$

$\prod\left(b^2+c^2\right)\ge$ $abc\prod (b+c)$ . See aici $\left\|\begin{array}{c} (a\cdot c+c\cdot b)^2\le \left(a^2+c^2\right)\left(c^2+b^2\right)\\\\ (a\cdot b+c\cdot c)^2\le \left(a^2+c^2\right)\left(b^2+c^2\right)\end{array}\right\|\bigodot\Longrightarrow$ $\left|c(a+b)\left(ab+c^2\right)\right|\le \left(a^2+c^2\right)\left(b^2+c^2\right)$ .

Since $x\le |x|$ obtain that $c(a+b)\left(ab+c^2\right)\le$ $\left(a^2+c^2\right)\left(b^2+c^2\right)$ with the equality $a=b=c$ .

Proof 2. $\left\{\begin{array}{ccc} 2(a^2+b^2) \geq (a+b)^2 & ; & a^2+b^2 \geq 2ab\\\\ 2(b^2+c^2) \geq (b+c)^2 & ; & b^2+c^2 \geq 2bc\\\\ 2(c^2+a^2) \geq (c+a)^2 & ; & c^2+a^2 \geq 2ca\end{array}\right\|\bigodot\implies$ $\left[\prod\left(b^2+c^2\right)\right]^2 \ge \left[abc\prod (b+c)\right]^2\iff$ $\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right) \geq abc(a+b)(b+c)(c+a)$ .

An easy extension. Prove that $\{m,n,p,a,b,c\}\subset\left[0,\infty\right)\Longrightarrow$ $\prod\left(b^2+c^2+mbc\right)\ge$ $abc\prod\left(b+c+\sqrt {npbc}\right)$ .

Proof. $\boxed{\left\{\begin{array}{ccc} \left(a^2+b^2+pab\right)\left(c^2+a^2+nca\right)\ge \left(ac+ba+a\sqrt{pncb}\right)^2=a^2\left(c+b+pn\sqrt{cb}\right)^2\\\\ \left(b^2+c^2+mbc\right)\left(a^2+b^2+pab\right)\ge \left(ba+cb+b\sqrt{mpac}\right)^2=b^2\left(a+c+mp\sqrt{ac}\right)^2\\\\ \left(c^2+a^2+nca\right)\left(b^2+c^2+nbc\right)\ge \left(cb+ac+c\sqrt{nmba}\right)^2=c^2\left(b+a+nm\sqrt{ba}\right)^2\end{array}\right\|\bigodot\implies\prod\left(b^2+c^2+mbc\right)\ge\ abc\prod\left(b+c+\sqrt {npbc}\right)}$

P12. Prove that in any triangle $ABC$ there are the inequalities $\left\{\begin{array}{ccc} \sum\sqrt{\frac a{b+c-a}} & \le & 1+\frac Rr\\\\ \sum\sqrt{\frac {a^3}{b+c-a}} & \le & \frac {Rs}r\end{array}\right\|$ .

Proof. I"ll use the well-known inequality $\boxed{\cos\frac {B-C}2\ge \sqrt{\frac {2r}R}}\ \odot 2\cos\frac {B+C}2=2\sin\frac A2\ \iff$ $2\cos\frac {B+C}2\cos\frac {B-C}2\ge 2\sin\frac A2\cdot \sqrt{\frac {2r}R}\iff$

$\cos B+\cos C\ge\sqrt {\frac {2r}R\cdot 4\sin^2\frac A2}=$ $\sqrt {\frac {2r}R\cdot \frac {4(s-b)(s-c)}{bc}}=$ $\sqrt {\frac {2r}R\cdot\frac {4(s-b)(s-c)}{bc}\cdot \frac {a(s-a)}{a(s-a)}}=$ $\sqrt {\frac {2r}R\cdot\frac {4sr^2}{4Rsr}\cdot \frac a{s-a}}=$ $\frac rR\sqrt{\frac {2a}{s-a}}=$ $\frac {2r}R\sqrt{\frac {a}{b+c-a}}\iff$

$\boxed{\cos B+\cos C\ge \frac {2r}R\sqrt{\frac {a}{b+c-a}}}\ (*)$ $\ \implies\ \sum \sqrt{\frac {a}{b+c-a}}\le$ $\frac R{2r}\cdot \sum (\cos B+\cos C)=$ $\frac Rr\cdot \sum \cos A=$ $\frac Rr\cdot \left(1+\frac rR\right)=$ $1+\frac Rr\iff$ $\boxed{\sum \sqrt{\frac {a}{b+c-a}}\le 1+\frac Rr}$ .

Otherwise. Prove easily that $\boxed{\sum (s-b)(s-c)=r(4R+r)}\ (1)$ and $\boxed{\sum a(s-b)(s-c)=2sr(2R-r)}\ (2)$ . Therefore,

$\sum (b+c)(s-b)(s-c)=$ $2s\sum (s-b)(s-c)-\sum a(s-b)(s-c)\ \stackrel{(1\wedge 2)}{\implies}\ 2sr(4R+r)$ $-2sr(2R-r)=4sr(R+r)=$

$\boxed{\sum (b+c)(s-b)(s-c)=4sr(R+r)}\ (3)$ . In conclusion, $\sqrt {s(b+c-a)}\le \frac {a+(b+c-a)}2=\frac {b+c}2\implies$ $\sqrt{\frac a{b+c-a}}=\frac {\sqrt{a(b+c-a)}} {b+c-a}\le \frac {b+c}{2(b+c-a)}\implies$

$\sum\sqrt{\frac a{b+c-a}}\le\sum \frac {b+c}{2(b+c-a)}=$ $\frac 14\sum\frac {b+c}{s-a}=$ $\frac 14\sum\frac {(b+c)(s-b)(s-c)}{(s-a)(s-b)(s-c)}\ \stackrel{(3)}{=}\ \sum\frac {4sr(R+r)}{4sr^2}=$ $\sum\frac {R+r}r\implies$ $\boxed{\sum\sqrt{\frac a{b+c-a}}\le 1+\frac Rr}$ .

$\blacktriangleright\ \sum \left[a(\cos B+\cos C)\right]\ge\frac {2r}R\cdot\sum\sqrt {\frac {a^3}{b+c-a}}\iff$ $a+b+c\ge\frac {2r}R\cdot\sum\sqrt {\frac {a^3}{b+c-a}}\iff$ $\boxed{\sum\sqrt{\frac {a^3}{b+c-a}}\le \frac {Rs}r}$ .

Remark. $\boxed{\cos\frac {B-C}2\ge \sqrt{\frac {2r}R}}\ \odot 2\sin\frac {B+C}2=2\cos\frac A2\ \iff$ $2\sin\frac {B+C}2\cos\frac {B-C}2\ge 2\cos\frac A2\cdot \sqrt{\frac {2r}R}\iff$ $\sin B+\sin C\ge\sqrt {\frac {2r}R\cdot4\cos^2\frac A2}\iff$

$\frac {b+c}{2R}\ge \sqrt{\frac {2r}R\cdot\frac {4s(s-a)}{bc}}\iff$ $bc(b+c)^2\ge 32Rsr(s-a)\iff$ $bc(b+c)^2\ge8abc(s-a)\iff$ $(b+c)^2\ge 4a(b+c-a)\iff$

$[(\underline{b+c-a})+\underline a]^2\ge 4\underline a(\underline {b+c-a})$ , what is truly. Have equality iff $b+c-a=a$ , i.e. $\boxed{b+c=2a}$ . See here

P13. Let $\{a,b,c,x,y,z\}\subset\mathbb R^*_+$ . Prove that $\frac x{ay+bz}+\frac y{az+bx}+ \frac z{ax+by}\ge \frac 3{a+b}$ .

Proof. $\frac {x^2}{axy+bzx}+\frac {y^2}{ayz+bxy}+\frac {z^2}{azx+byz}\ \stackrel{(C.B.S)}{\geq}\ \frac{(x+y+z)^2}{(a+b)(xy+yz+zx)}$ $\ge\frac3{a+b}\ ,$ which results from the well-known inequality $(x+y+z)^2 \geq 3(xy+yz+zx)$ .

P14. Ascertain $m\in\mathbb R$ so that the equation $\sin 2x=m(\sin x+\cos x)$ has at least a zero in $\left(0,\frac {\pi}2\right)$ .

Proof. Rewriting the equation as $\frac{2}{m}=\frac{1}{\sin x}+\frac{1}{\cos x} \geq \frac{4}{\sin x+\cos x} \geq \frac{4}{\sqrt{2}}$ . So, $m \leq \frac{1}{\sqrt{2}}$ . Also, $\frac{1}{m}=\frac{1}{2}\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right) \leq \sqrt{\frac{1}{2\sin^2x\cos^2x}}= \frac{\sqrt{2}}{\sin 2x}$ .

So, $m \geq 0$ (Its also obvious from the original equation itself). Also, we get solutions (easy to check) at both the boundary values of $m$ . So, $m\in\left[0,\frac{1}{\sqrt{2}}\right]$ .

P15. Prove that the inequality $\boxed {\ \frac {\cos^2A}{a^2} + \frac {\cos^2B}{b^2} + \frac {\cos^2C}{c^2}\ \ge\ \frac {1}{4R^2}\cdot \left(\frac {a^2 + b^2 + c^2}{4S\sqrt 3}\right)^2\ }\ \ge\ \frac {1}{4R^2}$ .

Proof. $4R^2\cdot\sum \frac {\cos^2A}{a^2} =$ $\sum\cot^2A = \sum\left(\frac {b^2 + c^2 - a^2}{4S}\right)^2\ge$ $\frac {1}{16S^2}\cdot\frac 13\cdot\left[\sum\left(b^2 + c^2 - a^2\right)\right]^2 = \left(\frac {a^2 + b^2 + c^2}{4S\sqrt 3}\right)^2\ge 1$ .

P16. Prove that $\{a,b,c,d\}\subset \mathbb R^*_+$ and $abc+abd+acd+bcd=4\implies$ $\left\{\begin{array}{ccc} a+b+c+d & \ge & 4\\\\ a^2+b^2+c^2+d^2 & \ge & 4\end{array}\right\|\ .$

Proof. $4=abc+abd+acd+bcd=$ $ab(c+d)+cd(a+b)\le$ $\left(\frac {a+b}2\right)^2\cdot (c+d)+\left(\frac {c+d}2\right)^2\cdot (a+b)=$ $(a+b)(c+d)\cdot \frac {\sum a}4\le$ $\left[\frac {(a+b)+(c+d)}2\right]^2\cdot \frac {\sum a}4=$

$\frac {(\sum a)^3}{16}\le$ $\frac {\left(2\sqrt{a^2+b^2+c^2+d^2}\right)^3}{16}=\frac 12\cdot \left(\sum a^2\right)^{\frac 32}$ $\implies$ $8\le \left(\sum a^2\right)^{\frac 32}\implies$ $8^{\frac 23}\le \sum a^2\implies 4\le a^2+b^2+c^2+d^2\ .$ Remark. $4\le \frac {(\sum a)^3}{16}\implies$ $\boxed{\sum a\ge 4}\ .$

P17. Sa se arate ca intr-un triunghi $ABC$ exista echivalenta $:$

$\boxed{\ \cos\ (B-C)\ \le\ \frac {2bc}{b^2+c^2}\ \le\ 1\ \Longleftrightarrow\ A\ \le\ 90^{\circ}\ \ \vee\ \ B=C\ \Longleftrightarrow\ \frac rR\ \le\ \frac {a(b+c-a)}{b^2+c^2}\ \le\ \frac 12\ \Longleftrightarrow\ \frac {2a(p-b)(p-c)}{p\left(b^2+c^2\right)}\ \le\ \frac rR\ }$ (demonstratia aici).

Proof. $\boxed{\cos (B-C)\le \frac {2bc}{b^2+c^2}\ }\ .$ Adunam $1$ stanga/dreapta si folosim $1+\cos x=2\cos^2\frac x2$ $\Longleftrightarrow$ $\boxed{2\cos^2\frac {B-C}{2}\le \frac {(b+c)^2}{b^2+c^2}}\ .$ Inmultim cu $2\sin ^2\frac {B+C}{2}=2\cos^2\frac A2>0\ .$

Folosim $2\sin\frac {B+C}{2}\cos\frac {B-C}{2}=$ $\sin B+\sin C$ si $\cos^2\frac A2=\frac {p(p-a)}{bc}$ $\Longleftrightarrow$ $\boxed{(\sin B+\sin C)^2\le \frac {(b+c)^2}{b^2+c^2}\cdot\frac {2p(p-a)}{bc}}\ .$ Insa $\sin B=\frac {b}{2R}\ ,$ $\sin C=\frac {c}{2R}$ $\Longleftrightarrow$

$\boxed{bc\left(b^2+c^2\right)\le 4R^2\cdot 2p(p-a)}\ .$ Inmultim cu $r$ si folosim $4Rpr=abc$ $\Longleftrightarrow$ $\boxed{rbc\left(b^2+c^2\right)\le R\cdot abc\cdot 2p(p-a)}\ .$ Simplificam prin $bc>0$ si stim ca $2(p-a)=b+c-a$

$\Longleftrightarrow$ $\boxed{\frac rR\le\frac {a(b+c-a)}{b^2+c^2}}$ $\Longleftrightarrow$ $\frac {4(p-a)(p-b)(p-c)}{abc}\le \frac {2a(p-a)}{b^2+c^2}$ $\Longleftrightarrow\ 2\left(b^2+c^2\right)(p-b)(p-c)\le a^2bc$ $\Longleftrightarrow$ $\left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\le 2a^2bc$ $\Longleftrightarrow$

$a^2\left[\left(b^2+c^2\right)-2bc\right]\le\left(b^2+c^2\right)(b-c)^2$ $\Longleftrightarrow$ $a^2(b-c)^2\le \left(b^2+c^2\right)(b-c)^2$ $\Longleftrightarrow$ $\boxed{B=C\ \ \vee\ \ A\ \le\ 90^{\circ}}$ deoarece $a^2\le b^2+c^2\Longleftrightarrow A\le 90^{\circ}\ .$

P18.Aratati ca in orice triunghi avem $OH \geq R|\sin(B-C)$ si $IH\ge\frac{b+c}{2p}\sqrt{4p(p-a)\left(\frac{bc}{(b+c)^2}-\frac{(p-b)(p-c)}{a^2}\right)}\ .$

Demonstratie 1. In triunghiul $AOH\ :\ \frac{AH}{|\sin (B-C)|}=\frac{R}{\sin AHO}\Longrightarrow OH=\frac{R|\sin (B-C)|}{\sin AHO}\ge R|\sin (B-C)|\ .$

Demonstratie 2. $OH\ \ge\ pr_{BC}OH=pr_{BC}OA=R\cdot \left|\cos\left(90^{\circ}-B+C\right)\right|=R\cdot \left|\sin (B-C)\right|\ .$ Procedam asemanator si pentru $IH\ :$

$IH\ \ge\ pr_{BC}IH=$ $pr_{BC}AI=\frac{b+c}{2p}\sqrt{b^2_a-h^2_a}\Longleftrightarrow$ $IH\ge\frac{b+c}{2p}\sqrt{4p(p-a)\left(\frac{bc}{(b+c)^2}-\frac{(p-b)(p-c)}{a^2}\right)}\ .$

P19. Sa se arate ca $:\ \boxed{\ \begin{array}{cc} A=60^{\circ}\ \Longleftrightarrow\ OI=2R\cdot\sin\frac {|B-C|}{4} & (1)\\\\ OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}4} & (2)\\\\ OI\ \ge\ 2R\cdot \sin\frac {|B-C|}{4}\sqrt {2\sin\frac A2} & (3)\end{array}\ }\ .$

Demonstratie.

$1\blacktriangleright$ $\boxed{\ \cos A+\cos B+\cos C=1+\frac rR\ }\ \wedge\ A=60^{\circ}\ \implies\ \cos B+\cos C=$ $\frac 12+\frac rR$ $\Longleftrightarrow$ $2\cos\frac{B+C}{2}\cdot\cos\frac {B-C}{2}=$ $\frac 12+\frac rR$ $\Longleftrightarrow$ $\cos\frac{B-C}{2}=\frac 12+\frac rR$ $\Longleftrightarrow$

$1-\cos\frac{B-C}{2}=$ $\frac 12-\frac rR\ \Longleftrightarrow$ $2sin^2\frac{B-C}{4}=\frac{R-2r}{2R}$ $\Longleftrightarrow$ $4R\sin^2\frac{B-C}{4}=R-2r$ $\Longleftrightarrow$ $4R^2\sin^2\frac{B-C}{4}=R^2-2Rr=OI^2$ $\Longleftrightarrow$ $\boxed{\ OI=2R\cdot\sin\frac{|B-C|}{4}\ }\ .$

$2\blacktriangleright$ Folosim $\cos A=1-2\sin^2\frac A2\ .$ Asadar, $\cos A+\cos B+\cos C=1+\frac rR\Longleftrightarrow$ $\cos B+\cos C=\frac rR+2\sin^2\frac A2$ $\Longleftrightarrow$ $2\cos\frac{B+C}2\cos\frac{B-C}2=$ $\frac rR+2\sin^2\frac A2$ $\Longleftrightarrow$

$2\sin\frac A2\cos\frac{B-C}{2}=\frac{r}{R}+2\sin^2\frac A2$ $\Longleftrightarrow$ $4R^2\sin\frac A2\left(1-2\sin^2\frac{B-C}4\right)=$ $2Rr+4R^2\sin^2\frac A2$ $\Longrightarrow$ $OI^2=R^2-2Rr=$ $R^2+4R^2\sin^2\frac A2-4R^2\sin\frac A2+$

$8R^2\sin\frac A2\sin^2\frac{B-C}{4}$ $\Longleftrightarrow$ $OI^2=R^2\left(2\sin^2\frac A2-1\right)^2+8R^2\sin\frac A2\sin^2\frac{B-C}{4}$ $\Longleftrightarrow OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}}\ .$

$3\blacktriangleright$ Inlocuind pe $OI$ si ridicand la patrat, inegalitatea de demonstrat devine $:\ \left(2\sin\frac A2-1\right)^2\ge 0\ ,$ ceea ce este adevarat. Egalitatea are loc daca si numai daca $A=60^{\circ}\ .$

P20. Aratati ca in orice $\triangle ABC$ are loc inegalitatea $:\ \sum \frac{a}{AI^2} \geq \frac{p}{2r^2}\ .$

Demonstratie 1. $\mathrm{LHS}=\frac{1}{4R}\sum \frac{\sin A}{\sin^2\frac{B}{2}\cdot\sin^2\frac{C}{2}}\ge\frac{1}{4R}\frac{\prod\cos\frac{A}{2}}{\prod\sin^2\frac{A}{2}}=\mathrm{RHS}\ .$ Asadar, $\sum \sin A\ge\sum \sin 2A\Longleftrightarrow$ $\prod \sin \frac{A}{2}\le \frac{1}{8}\ .$ Altfel. $\sum \frac{a}{AI^2}=\frac{p(R-r)}{r^2R}\geq \frac{p}{2r^2}\ .$

Demonstratie 2. Vom demonstra fara a folosi identitatea $\boxed{\ \sum a^2(p-b)(p-c)=4p^2r(R-r)\ }\ .$ Astfel, $IA^2=\frac {bc(p-a)}{p}$ $\Longrightarrow\sum\frac {a}{IA^2}=\sum\frac {ap}{bc(p-a)}=\frac {p}{abc}\sum\frac {a^2}{p-a}=$

$\frac {1}{4Rr}\sum\frac {a^2}{p-a}\ .$ Insa $\sum\frac {a^2}{p-a}\stackrel{C.B.S.}{\ \ \ge\ \ }\frac {\left(\sum a\right)^2}{\sum (p-a)}=4p\ .$ Asadar $\sum\frac {a}{IA^2}\ge \frac {p}{Rr}\ ,$ insa mai slaba decat cea propusa deoarece $\frac {p}{2r^2}\ge\frac {p}{Rr}\ .$

Consecinta. Sa se arate ca in orice $\triangle ABC$ exista relatia $\boxed{\ \max\left\{\ \frac {a}{p-a}\ ,\ \frac {b}{p-b}\ ,\ \frac {c}{p-c}\ \right\}\ \ge \ \frac Rr\ }\ .$ Indicatie "tare" $:\ \frac {a\cdot\frac {a}{p-a}+b\cdot\frac {b}{p-b}+c\cdot\frac {c}{p-c}}{a+b+c}\ \ge\ \frac {R}{r}\ .$

P21. Sa se arate ca daca $\triangle ABC$ este ascutitunghic atunci $2[p^2-R(R-2r)]\ \le\boxed{\sum\frac {a(a^2+2bc)}{b+c}\ \le\ \frac {4(R+r)^2(2R-r)}{R}\ \le\ 9R(2R-r)}\ .$

Proof. Prima inegalitate este adevarata in orice $\triangle ABC$ (demonstrate !). Revenim. Folosind teorema Cosinusului partea stanga a inegalitatii este echivalenta cu $:\ \sum\ \frac{a(a^2+2bc)}{b+c}=$

$\sum\ \frac{a(b^2+c^2-2bc\cos A+2bc)}{b+c}=$ $\sum\ \frac{a[(b+c)^2-2bc\cos A]}{b+c}=$ $\sum\ a(b+c)-2abc\sum\ \frac{\cos A}{b+c}\ .$ Asadar, $LHS\equiv 2(ab+bc+ca)-2abc\cdot \left(\sum \frac{\cos A}{b+c}\right)\ (\ast)\ .$

In continuare vom demonstra inegalitatea (valabila in triunghi ascutitunghic !) $:\ \boxed{\sum\ \frac{\cos A}{b+c}\ \ge\ \frac{(R+r)^2}{2R^2p}}\ .$ Intr-adevar, aplicand inegalitatea C.B.S. si folosind faptul ca

$\boxed{\sum\ \cos A=1+\frac rR}$ si $\boxed{a=b\cos C+c\cos B}$ obtinem $:\ \sum\ \frac{\cos A}{b+c}=$ $\sum\frac{\cos^2 A}{(b+c)\cos A}\ \stackrel{\mathrm{C.B.S}}{ge}\ \frac{\left(\sum\cos A\right)^2}{\sum\ (b\cos A+c\cos A)}=$ $\frac{\left(1+\frac rR\right)^2}{\sum\ (b\cos C+c\cos B)}=$ $\frac{\frac{(R+r)^2}{R^2}}{\sum\ a}=$

$\frac{(R+r)^2}{2R^2p}\ .$ Revenind in $\stackrel{(*)}{\iff}\ LHS\le 2(ab+bc+ca)-8Rrp\ \cdot\ \frac{(R+r)^2}{2R^2p}=$ $2(ab+bc+ca)-\frac{4r(R+r)^2}{R}\ .$ In continuare aplicam inegalitatea Gerretsen $:$

$\boxed{ab+bc+ca\le 4(R+r)^2}\ \iff\ LHS\ \le\ 8(R+r)^2-\frac{4r(R+r)^2}{R}\iff$ $\boxed{\sum\ \frac{a(a^2+2bc)}{b+c}\ \le\ \frac{4(R+r)^2(2R-r)}R}\ .$ Frumoasa inegalitate !

P22. Sa se arate ca in $\triangle ABC$ cu cercul circumscris $w$ exista relatiile $\boxed{2(a+b+c)\le\sum\frac {bc}{s-a}\le\frac {(5R-2r)(4R+r)^2}{2s(2R-r)}}$ (notatii standard).

Demonstratie. Vom folosi relatia metrica remarcabila $\boxed{H\Gamma^2=4R^2\left[1-\frac{2s^2(2R-r)}{R(4R+r)^2}\right]\ge 0}\ (**)\ ,$ unde $H$ este ortocentrul si $\Gamma$ este punctul lui Gergonne pentru $\triangle ABC$ de unde

rezulta usor $\boxed{2s^2(2R-r)\le R(4R+r)^2}\ (*)\ .$ Asadar, $\sum\frac {bc}{s-a}=\frac {abc}s\cdot\sum\left(\frac 1{s-a}+\frac 1a\right)=$ $4Rr\cdot\left(\sum\frac 1{s-a}+\sum\frac 1a\right)=$ $4Rr\cdot \left[\frac {r(4R+r)}{sr^2}+\frac {s^2+r(4R+r)}{4Rrs}\right]=$

$\frac {4R(4R+r)}s+s+\frac {r(4R+r)}s=$ $s+\frac {(4R+r)^2}s\implies$ $\boxed{\sum\frac {bc}{s-a}=s+\frac {(4R+r)^2}s}\ (1)\ .$ In concluzie, inegalitatea propusa este echivalenta cu inegalitatea $s+\frac {(4R+r)^2}s\le$

$\frac {(5R-2r)(4R+r)^2}{2s(2R-r)}\iff$ $2s^2(2R-r)+2(2R-r)(4R+r)^2\le (5R-2r)(4R+r)^2\iff$ $2s^2(2R-r)\le \left[(5R-2r)-(4R-2r)\right](4R+r)^2\iff$

$2s^2(2R-r)\le R(4R+r)^2$ , adica ineg. adevarata $(*)\ .$ Se observa ca $\sum\frac {bc}{s-a}\ge 4s\iff$ $\frac {(4R+r)^2}s\ge 3s\iff$ $s\sqrt 3\le 4R+r$ , care este cunoscuta a fi adevarata.

Observatie. Vom dovedi relatia remarcabila $(**)$ folosind identitatea de tip Leibniz $\boxed{\sum (s-b)(s-c)\cdot HA^2=r(4R+r)\cdot H\Gamma ^2+\frac{4s^2r^2(R+r)}{4R+r}}\ (2)$ unde $\Gamma$ are coordonatele

$\left(\frac 1{s-a},\frac 1{s-b},\frac 1{s-c}\right)$ in raport cu $w\ .$ Se stie ca $\sum (s-b)(s-c)=r(4R+r)$ si puterea lui $\Gamma$ fata de $w$ este $p_w\left(\Gamma\right)=$ $-r(R+r)\cdot\left(\frac {2s}{4R+r}\right)^2\ .$ Se arata usor relatia

$\boxed{\sum a^2(s-b)(s-c)=4s^2r(R-r)}\ (***)\ .$ Asadar, $HA=2R|\cos A|\implies$ $HA^2=4R^2-a^2$ $\implies$ $\sum (s-b)(s-c)\cdot HA^2=$ $\sum (s-b)(s-c)\cdot \left(4R^2-a^2\right)=$

$4R^2\sum (s-b)(s-c)-$ $\sum a^2(s-b)(s-c)\ \stackrel{(***)}{=}\ 4R^2r(4R+r)-4s^2r(R-r)$ $\implies$ $\boxed{\sum (s-b)(s-c)\cdot HA^2=4R^2r(4R+r)-4s^2r(R-r)}\ (3)\ .$ Din relatia

Leibniz $(2)$ si relatia precedenta $(3)$ obtinem $4R^2r(4R+r)-4s^2r(R-r)=$ $r(4R+r)\cdot H\Gamma ^2+$ $\frac{4s^2r^2(R+r)}{4R+r}$ $\iff$ $H\Gamma^2=4R^2-\frac{4s^2(R-r)}{4R+r}-\frac{4s^2r(R+r)}{(4R+r)^2}$ $\iff$

$H\Gamma^2=4R^2-\frac{4s^2}{(4R+r)^2}\cdot[(4R+r)(R-r)+r(R+r)]=$ $4R^2-\frac{4s^2}{(4R+r)^2}\cdot 2R(2R-r)=$ $4R^2-\frac {8s^2R(2R-r)}{(4R+r)^2}\implies$ $H\Gamma^2=4R^2\left[1-\frac{2s^2(2R-r)}{R(4R+r)^2}\right]\ .$

P23. Let $\triangle ABC$ with the incircle $C(I)$ and the circumcircle $w=\mathbb C(O,R)\ .$ Consider $D\in (BC)\ ,$ $E\in (CA)\ ,$ $F\in (AB)$ and denote $\{A,X\}=AD\cap w\ ,$

$\{B,Y\}=BE\cap w\ ,$ $\{C,Z\}=CF\cap w\ .$ Prove that $\frac {XA}{XD}+\frac {YB}{YE}+\frac {ZC}{ZF}\ \ge\ \left(\frac {b+c}{a}\right)^2+\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2$ with equality iff $I\ \in\ AD\cap BE\cap CF\ .$

Proof. I"ll prove $:\ \frac{XA}{XD}\ \ge\ \left(\frac {b+c}a\right)^2\ ,$ with equality iff $AD$ is the bisector of $\widehat {BAC}\ .$ Apply the Stewart's relation to the cevian $AD/\triangle ABC\ :\ a\cdot AD^2+a\cdot BD\cdot CD=$

$c^2\cdot CD+b^2\cdot BD$ $\Longrightarrow \frac{AD^2}{BD\cdot CD} + 1= \frac{c^2}{a\cdot BD}+\frac{b^2}{a\cdot CD}\ \stackrel{\small C.B.S.}{\ge}\ \left(\frac{b+c}a\right)^2$ cu egalitate iff $\frac{c}{a\cdot BD}=\frac{b}{a\cdot CD}\iff$ $AD$ is the bisector of $\widehat{BAC}\ .$ But $X\in AD\cap w\ ,$

$A\ne X\ \Longrightarrow$ $BD\cdot CD=AD\cdot DX\ .$ The last inequality is equivalent with $\frac{AD^2+AD\cdot DX}{AD\cdot DX}\ \ge\$ $\left(\frac {b+c}a\right)^2\ \Longleftrightarrow\$ $\frac{XA}{XD}\ \ge\ \left(\frac {b+c}a\right)^2$ $\implies$ $\sum \frac{XA}{XD}\ \ge\ \sum \left(\frac {b+c}a\right)^2\ .$

P24. Fie $\{a,b,c,d\}\subset R^*$ so that $\left\{\begin{array}{ccc} a+b+c+d & = & 10\\\\ a^2+b^2+c^2+d^2 & = & 28\end{array}\right\|\ .$ Prove that $abcd\ne 0\ ,$ $d\le 4$ and find the maximum of the sum $\sum\frac 1a\ .$

Proof. $\left\{\begin{array}{c} a+b+c=10-d\\\\ a^2+b^2+c^2=28-d^2\\\\ (a+b+c)^2\le 3\left(a^2+b^2+c^2\right)\implies\end{array}\right\|\implies$ $(10-d)^2\le 3\left(28-d^2\right)\implies$ $d^2-5d+4\le 0\implies$ $\boxed{\ \{a,b,c,d\}\subset [1,4]\ }\ .$

Apply the P. Schweitzer's inequality $\boxed{x_k\in [a,b]\ ,\ k\in\overline {1,n}\implies n^2\le \sum x_k\cdot\sum\frac {1}{x_k}\le \frac {(a+b)^2}{4ab}\cdot n^2}$ and obtain that $\sum \frac 1a\in \left[\frac 85,\frac 52 \right]\ .$

P25. Sa se arte ca in orice triunghi $ABC$ exista inegalitatea $a^2+b^2+c^2\ge 4S\cdot \left(\tan\frac A2+\tan\frac B2+\tan\frac C2\right)+\frac {8r}{R}\cdot OI^2\iff$ $s^2+5r^2\ge 16Rr\ .$

Demonstratie. Se folosesc cunoscutele identitati geometrice $\sum r_a=4R+r\ ,$ $\tan\frac A2=\frac r{s-a}$ si $OI^2=R(R-2r)$ (notatii standard).

Intr-adevar, $\sum a^2\ge 4S\cdot\sum\tan \frac A2+\frac {8r}R\cdot OI^2\iff$ $2\left(s^2-r^2-4Rr\right)\ge 4S\sum\frac r{s-a}+\frac {8r}R\cdot\left(R^2-2Rr\right)\iff$

$s^2\ge r^2+4Rr+2r\sum r_a+4r(R-2r)\iff$ $s^2\ge 8Rr-7r^2+2r(4R+r)\iff$ $s^2+5r^2\ge 16Rr\ ,$ what is true.

P26. Let $\{x,y\}\subset\mathbb R_+^*$ so that $xy<1\ .$ Prove that $\boxed{\left(\frac{2x}{1+x^2}\right)^2+\left(\frac{2y}{1+y^2}\right)^2\le\frac {1}{1-xy}}\ (*)\ .$

Proof. $(\exists )\ \{a,b\}\subset \left(0,\frac {\pi}2\right)$ so that $\left\{\begin{array}{ccc} x=\tan a\\\\ y=\tan b\end{array}\right\|$ and $xy<1\iff \tan a\tan b<1\iff$ $\sin a\sin b<\cos a\cos b\iff$ $\cos(a+b)>0\iff$ $a+b\in\left(0,\frac {\pi}2\right)\ .$

Our inequality is equivalent with $\sin^22a+\sin^22b\le \frac 1{1-\tan a\tan b}\iff$ $E\ge 0$ where $E\equiv \tan a\tan b\left(\sin^2 2a+\sin^2 2b\right)+1-\sin^2 2a-\sin^2 2b\ .$ Observe

that $E\ge 2\tan a\tan b\sin 2a\sin 2b+1-\sin^2 2a-\sin^2 2b=$ $8\sin^2a\cdot\sin^2b+1-4\sin^2 a\left(1-\sin^2 a\right)-4\sin^2 b\left(1-\sin^2b\right)=$ $4\left(\sin^2a+\sin^2b\right)^2+$

$1-4\left(\sin^2a+\sin^2b\right)=$ $\left(2\sin^2 a+2\sin^2b-1\right)^2\ge 0\ .$ Equality occurs for $:$ if $a,b>0$ then $\sin 2a=\sin 2b,\;\sin^2a+\sin^2b=\frac{1}{2}\iff$ $a=b\ ,\ \sin a=\frac{1}{2}\ ,$

$\{a,b\}\subset\left[0,\frac{\pi}4\right]\iff$ $a=\frac{\pi}6\ ,\ b=\frac{\pi}6\iff$ $x=y=\frac{1}{\sqrt{3}}\ ;$ if $a=0$ or $b=0$ then $b=\frac{\pi}{4}$ or $a=\frac{\pi}{4}\iff$ $x=0\ ,\ y=1$ or $x=1\ ,\ y=0\ .$

P27. Let $\triangle ABC$ and $\{\alpha , \beta , \gamma\}\subset\mathbb R\ .$ Prove that $\left\{\begin{array}{cccccccc} \alpha +\beta +\gamma & = & 0 & \implies & \beta\gamma a^2+\gamma\alpha b^2+\alpha \beta c^2 & \le & 0 & (1)\\\\ \alpha a+\beta b+\gamma c & = & 0 & \implies & \alpha\beta +\beta \gamma +\gamma\alpha & \le & 0 & (2)\end{array}\right\|\ .$

Proof.

$1.\blacktriangleright$ Suppose w.l.o.g. $\alpha\beta\gamma\ne 0\ .$ Thus, $\beta\gamma a^2+\gamma\alpha b^2+\alpha \beta c^2\le 0\iff$ $\beta\gamma a^2\le -\alpha \left(\gamma b^2+\beta c^2\right)\iff$ $\beta\gamma a^2\le (\beta +\gamma )\left(\gamma b^2+\beta c^2\right)\iff$

$\beta^2c^2+\beta\gamma \left(b^2+c^2-a^2\right)+\gamma^2b^2\ge 0\iff$ $\beta^2c^2+2\beta\gamma bc\cos A+\gamma^2b^2\ge 0\iff$ $\left(\beta c+\gamma b\cos A\right)^2+\gamma^2b^2\sin^2A\ge 0\ ,$ what is true.

We have equality, i.e. $\beta\gamma a^2+\gamma\alpha b^2+\alpha \beta c^2= 0\iff\alpha =\beta =\gamma=0\ .$

$2.\blacktriangleright$ Suppose w.l.o.g. $\alpha\beta\gamma\ne 0$ and denote $\boxed{t=\frac {\beta}{\gamma}}\ (*)\ .$ Thus, $-\alpha =\frac {\beta b+\gamma c}a$ and $-(\alpha\beta +\beta \gamma +\gamma\alpha) =$ $-\beta\gamma -\alpha (\beta +\gamma)=$ $-\beta\gamma +\frac {(\beta +\gamma)(\beta b+\gamma c)}a\ .s.s.$

$-a\beta\gamma +$ $(\beta +\gamma)(\beta b+\gamma c)\ .s.s.\ -at +$ $(t+1)(bt+c)=$ $bt^2+(b+c-a)t+c\ .$ Denote $\boxed{f(t)=bt^2+(b+c-a)t+c}\ .$ Observe that $\Delta =(b+c-a)^2-4bc=$

$(b+c)^2-4bc+a^2-2a(b+c)=$ $(b-c)^2+a^2-2a(b+c)\le$ $a^2+a^2-2a(b+c)=$ $2a^2-2a(b+c)\le 0\implies$ $\Delta\le 0\ ,$ i.e. $f(t)\ge 0$ for any $t\in\mathbb R^*\ .$

In conclusion, $-(\alpha\beta +\beta \gamma +\gamma\alpha)\ge 0\iff \alpha\beta +\beta \gamma +\gamma\alpha\le 0\ .$

Application. Let $\triangle ABC$ and $M\in (BC)\ ,$ $N\in (CA)\ ,$ $P\in (AB)$ so that $AP+BM+CN=PB+MC+NA\ .$ Prove that $[MNP]\le \frac S4\ ,$ where $S=[ABC]\ .$

Proof. Denote $AP=\gamma c\ ,$ $BM=\alpha a\ ,$ $CN=\beta b\ ,$ i.e. $BP=(1-\gamma ) c\ ,$ $CM=(1-\alpha ) a\ ,$ $BN=(1-\beta )b\ ,$ where $\{\alpha ,\beta ,\gamma\}\subset(0,1)\ .$ Thus, $AP+BM+CN=$

$PB+MC+NA\iff$ $\boxed{(2\alpha -1)a+(2\beta -1)b+(2\gamma -1)c=0}\ (*)\ .$ From the previous PP27 (the second implication) obtain $(2\alpha -1)(2\beta -1)+(2\beta -1)(2\gamma -1)+$

$(2\gamma -1)(\alpha -1)\le 0$ $\iff$ $4\cdot \sum \beta\gamma-2\cdot\sum (\beta +\gamma )+3\le 0\iff$ $4\cdot \sum \beta\gamma-4\cdot\sum \alpha +3\le 0\iff$ $\boxed{\sum \alpha -\sum \beta\gamma\ge\frac 34}\ (2)\ .$ Prove easily that $\frac {[MNP]}{[ABC]} =$

$1-\frac {[PAN]}{[BAC]}-\frac {[MBP]}{[CBA]}-\frac {[NCM]}{[ACB]}=$ $1-\gamma (1-\beta )-\alpha (1-\gamma )-\beta (1-\alpha )=$ $1-\sum \alpha +\sum\beta\gamma\ \stackrel{(2)}{\le}\ 1-\frac 34=\frac 14\ ,$ i.e. $[MNP]\le\frac S4\ .$

P28 (own). $\triangle ABC\Longrightarrow \left(\forall\right)\ X\ ,\ \frac{XA^2}{a}+\frac{XB^2}{b}+\frac{XC^2}{c}\ge \frac{a^3+b^3+c^3}{ab+bc+ca}$ . Remark. Try some particular cases: $X\in \{O,H,G,\ldots\}\ .$

Proof (Constantin MATEESCU). The solution has been inspired from one of the problems in Virgil's book on Planar Geometry (este singura carte pe care o port cu mine oriunde!

). I don't think there

is a different way to derive this amazing inequality which features a very special equality case! Firstly, we need to recall that the distance between two points $P$ and $P^{\prime}$ whose normalised barycentric

coordinates w.r.t. $\triangle ABC$ are $(\alpha, \beta, \gamma)$ and $(\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime})$ respectively, is given by the following relation $:\ \boxed{ {PP^{\prime}}^2 = - \left[(\beta - \beta^{\prime}) (\gamma - \gamma^{\prime}) a^2 + (\gamma - \gamma^{\prime}) (\alpha - \alpha^{\prime}) b^2 + (\alpha - \alpha^{\prime}) (\beta - \beta^{\prime}) c^2\right] }\ .$

Consequence. For $P(\alpha, \beta, \gamma)$ and $P^{\prime}(\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime})$ w.r.t. $\triangle ABC$ the following inequality holds $:\ \boxed{ (\beta - \beta^{\prime}) (\gamma - \gamma^{\prime}) a^2 + (\gamma - \gamma^{\prime}) (\alpha - \alpha^{\prime}) b^2 + (\alpha - \alpha^{\prime}) (\beta - \beta^{\prime}) c^2 \le 0 }\ .$ Now let $P = X$

and let $P^{\prime}$ be the isotomic of the incenter $I$ of $\triangle ABC$ . Thus: $X(\alpha, \beta, \gamma )$ and $P^{\prime}\left(\frac {bc}{ab + bc + ca}, \frac {ca}{ab + bc + ca}, \frac {ab}{ab + bc +ca}\right)\ .$ Using the above corollary we succesively obtain that

$\sum a^2\left(\beta - \frac {ca}{ab + bc + ca}\right)\left(\gamma - \frac {ab}{ab + bc + ca}\right) \le 0\iff \sum a^2\beta\gamma - \sum \frac{a^3(b\beta + c\gamma)}{ab + bc + ca} + abc\sum \frac {a^3}{(ab + bc + ca)^2} \le 0\iff$ $\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \frac{a^3(b\beta + c\gamma)}{abc} -$

$\frac {ab + bc + ca}{abc}\sum a^2\beta\gamma\iff \frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \frac {bc(b^2\gamma + c^2\beta)}{abc} -$ $\frac {ab + bc + ca}{abc}\sum a^2\beta\gamma\iff$ $\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \left(\frac {bc(b^2\gamma + c^2\beta)}{abc} - \frac {bc}{abc}\sum a^2\beta\gamma \right)\iff$

$\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \frac {\left(b^2\gamma + c^2\beta\right) - \left(a^2\beta\gamma + b^2\gamma\alpha + c^2\alpha\beta\right)}a\iff$ $\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \frac {-\left[\beta\gamma a^2 + \gamma\left(\alpha - 1\right) b^2 + \left(\alpha - 1\right)\beta c^2\right]}a\iff$ $\boxed{\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \frac {XA^2}a + \frac {XB^2}b + \frac {XC^2}c}\ .$

The equality is attained $\iff\ X$ is the isotomic of the incenter $I$ of $\triangle ABC$ . Particular cases. $\boxed{\begin{array}{ccc} X:= H\ \mathrm{\left(\underline{orthocenter}\right)} & \implies & \sum\frac a{\tan^2A}\ge \frac {a^3+b^3+c^3}{ab+bc+ca}\\\\ X:=O\ \mathrm{\left(\underline{circumcenter}\right)} & \implies & \sqrt {abc\left(a^3+b^3+c^3\right)}\le R\cdot (ab+bc+ca)\end{array} }\ .$

P29 (Nguyen Viet Hung, Vietnam) <= click.

Proof 1 (with barycentric coordinates). $G\left(\frac 13,\frac 13,\frac 13\right)\ ,$ $I\left(\frac {a}{\sum a},\frac {b}{\sum a},\frac {c}{\sum a}\right)\ ,$ $K\left(\frac {a^2}{\sum a^2},\frac {b^2}{\sum a^2},\frac {c^2}{\sum a^2}\right)$ $\implies$ $\frac {[GIK]}{[ABC]}=\left|\begin{array}{ccc} \frac 13 & \frac 13 & \frac 13\\\\ \frac {a}{\sum a} & \frac {b}{\sum a} & \frac {c}{\sum a}\\\\ \frac {a^2}{\sum a^2} & \frac {b^2}{\sum a^2} & \frac {c^2}{\sum a^2}\end{array}\right|=$

$\frac 13\cdot \frac 1{\sum a}\cdot\frac 1{\sum a^2}\cdot \mathrm{mod}\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|=$ $\frac{\left|\ V(a,b,c)\ \right|}{3(a+b+c)\left(a^2+b^2+c^2\right)}=$ $\frac {|(c-a)(c-b)(b-a)|}{3(a+b+c)\left(a^2+b^2+c^2\right)}\implies$ $\boxed{\ \frac {[GIK]}{[ABC]}=\frac {|(c-a)(c-b)(b-a)|}{3(a+b+c)\left(a^2+b^2+c^2\right)}\ }\ .$

This post has been edited 62 times. Last edited by Virgil Nicula, Jan 21, 2018, 2:14 PM

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462. Diverse probleme de geometrie.

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162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

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155. A constant ratio in a triangle ABC.

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148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

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133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

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121. Viete's relations with cos A , cos B , cos C.

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106. OLM - Bucuresti (geometrie).

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88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

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60. Mediteranean M.C. 2010 Problem 3.

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57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

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44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

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24. A fine and cool inequalities.

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22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

457. Relatii metrice si inegalitati geometrice.

by Virgil Nicula, Aug 19, 2017, 11:10 AM

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