307. Very nice proofs !

by Virgil Nicula, Jul 31, 2011, 11:50 PM

$\blacktriangleright$ PP1. Prove the well-known identity $\boxed{ab+bc+ca=s^2+r^2+4Rr}$ , where $2s=a+b+c$ .

Proof 1. $(s-a)(s-b)(s-c)=sr^2\iff$ $s^3-2s^3+s\sum bc-4Rsr=sr^2\iff$ $ab+bc+ca=s^2+r^2+4Rr$ .

Commentary. I used the Viete's relations and the identities $S=pr=\sqrt {s(s-a)(s-b)(s-c)}\ ,\ abc=4RS=4Rsr$ .

Proof 2. Prove easily that $ABI_a\sim AIC$ from where obtain $AI\cdot AI_a=bc\ (1)$ . I"ll use and evident relation $IA^2=(s-a)^2+r^2\ (2)$ .

In conclusion, $ab+bc+ca=bc+a(b+c)=$ $bc+s^2-(s-a)^2\stackrel{(2)}{=}$ $bc+s^2-\left(IA^2-r^2\right)=$ $s^2+r^2+\left(bc-IA^2\right)\stackrel{(1)}{=}$

$s^2+r^2+\left(AI\cdot AI_a-IA^2\right)=$ $s^2+r^2+AI\cdot\left (AI_a-AI\right)=$ $s^2+r^2+2\cdot IA\cdot IA^{\prime}=$ $s^2+r^2+4Rr$ .

Commentary. Without of the relations $(1)$ si $(2)$ I used and the relation $IA\cdot IA^{\prime}=2Rr$ - the power of $I$ w.r.t. the circumcircle $w$ of $\triangle ABC$ ,

where $\{A,A^{\prime}\}=AI\cap w$ . Can use (see the following proof) and well-known identity $s(s-a)+(s-b)(s-c)=bc$ a.s.o.

Proof 3. $\sum bc=\sum [s(s-a)+(s-b)(s-c)]\implies$ $\boxed{\sum bc=s^2+\sum (s-b)(s-c)}$ and $4Rsr=abc=\prod [s-(s-a)]=$

$s^3-s^3+s\cdot \sum (s-b)(s-c)-\prod (s-a)=$ $s\cdot\sum (s-b)(s-c)-sr^2\implies$ $\boxed{\sum (s-b)(s-c)=r^2+4Rr}$ .

Remark. $\sum bc=s^2+r^2+4Rr\iff$ $\sum (s-b)(s-c)=$ $r\sum r_a=$ $r(4R+r)\iff$ $\sum a(s-a)=2r(4R+r)$ .

$\blacktriangleright$ PP2. $\triangle ABC\ \implies\ \boxed{4S\sqrt 3\le a^2+b^2+c^2}\le 9R^2$ .

Proof. Using the well-known inequality $3(xy+yz+zx)\le (x+y+z)^2$ and the identity $16S^2=$ $\sum\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)$

obtain for $\left\{\begin{array}{c} x:=b^2+c^2-a^2\\\ y:=c^2+a^2-b^2\\\ z:=a^2+b^2-c^2\end{array}\right\|$ that $3\cdot 16S^2=3\sum yz\le (x+y+z)^2=\left(a^2+b^2+c^2\right)^2$ $\implies$ $4S\sqrt 3\le a^2+b^2+c^2\le 9R^2$ .

Remark. $12r(2R-r)\le a^2+b^2+c^2\le 8R^2+4r^2$ .

$\blacktriangleright$ PP3. $\triangle ABC\ ,\ A\le 90^{\circ}\ \implies\ \boxed{\cos (B-C)\le\frac {2bc}{b^2+c^2}}\le 1$ .

Proof 1 (Mudok). $a^2=\left(b\cos C+c\cos B\right)^2\le \left(b^2+c^2\right)\left(\cos ^2B+\cos^2C\right)\iff$ $2a^2\le \left(b^2+c^2\right)\left(2+\cos 2A+\cos 2B\right)\iff$

$a^2\le \left(b^2+c^2\right)\left[1-\cos A\cos (B-C)\right]\iff$ $\left(b^2+c^2\right)\cos A\cos (B-C)\le b^2+c^2-a^2=2bc\cos A\iff$ $\cos (B-C)\le\frac {2bc}{b^2+c^2}$ .

Proof 2 (own). $(b-c)^2\cdot\cos A\ge 0\iff$ $\left(b^2+c^2\right)\cos A\ge 2bc\cdot\cos A\iff$ $\cos A\ge\frac {b^2+c^2-a^2}{b^2+c^2}\iff$ $1-\cos A\le\frac {a^2}{b^2+c^2}\iff$

$2\sin^2\frac A2\le\frac {a^2}{b^2+c^2}\iff$ $\boxed{\sin\frac A2\le\frac {a}{\sqrt{2\left(b^2+c^2\right)}}}\le 1$ . Using the well-known identity $\boxed{\ \cos\frac {B-C}{2}=\frac {b+c}{a}\cdot\sin\frac A2\ }$ obtain that

$\boxed{\cos\frac {B-C}{2}\le \frac {b+c}{\sqrt {2\left(b^2+c^2\right)}}}\le 1\iff$ $2\cos^2\frac {B-C}{2}\le\frac {(b+c)^2}{b^2+c^2}\iff$ $1+\cos (B-C)\le\frac {(b+c)^2}{b^2+c^2}\iff$ $\cos (B-C)\le\frac {2bc}{b^2+c^2}$ .

Proof 3 (metric - own). $\cos (B-C)=\cos B\cos C+\sin B\sin C=$ $\frac {a^2+c^2-b^2}{2ac}\cdot\frac {a^2+b^2-c^2}{2ab}+\frac {2S}{ac}\cdot\frac {2S}{ab}=$

$\frac {\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)+16S^2}{4a^2bc}=$ $\frac {2\left[a^4-\left(b^2-c^2\right)^2\right]+2a^2\left(b^2+c^2-a^2\right)}{4a^2bc}\implies$ $\boxed{\cos (B-C)=\frac {a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)^2}{2a^2bc}}$ .

In conclusion, $\cos (B-C)\le\frac {2bc}{b^2+c^2}\iff$ $\frac {a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)^2}{2a^2bc}\le\frac {2bc}{b^2+c^2}\iff$ $a^2\left(b^2+c^2\right)^2-\left(b^2+c^2\right)\left(b^2-c^2\right)^2\le 4a^2b^2c^2\iff$

$a^2\left[\left(b^2+c^2\right)^2-4b^2c^2\right]\le\left(b^2+c^2\right)\left(b^2-c^2\right)^2\iff$ $a^2\left(b^2-c^2\right)^2\le\left(b^2+c^2\right)\left(b^2-c^2\right)^2\iff$ $\left(b^2-c^2\right)^2\left(b^2+c^2-a^2\right)\ge 0$ , what is truly.

Proof 4 (trigonometric - own). $\cos (B-C)\le\frac {2bc}{b^2+c^2}\iff$ $\cos (B-C)\le\frac {2\sin B\sin C}{\sin^2B+\sin^2C}\iff$

$\cos (B-C)\left(\sin^2B+\sin^2C\right)\le\cos (B-C)+\cos A\iff$ $\cos (B-C)(\cos 2B+\cos 2C)+2\cos A\ge 0\iff$

$\cos (B-C)\cos (B+C)\cos (B-C)+\cos A\ge 0\iff$ $\cos A\left[1-\cos^2(B-C)\right]\ge 0$ , what is truly.

Proof 5 (own). Add $1$ left/right to the required relation $\cos (B - C)\le \frac {2bc}{b^2 + c^2}$ and I"ll use the relation $1 + \cos x = 2\cos^2\frac x2$ $\implies$ $\left\|\ 2\cos^2\frac {B - C}2\le \frac {(b + c)^2}{b^2 + c^2}\ \right\|$ .

Multiply by $2\sin ^2\frac {B + C}{2} = 2\cos^2\frac A2 > 0$ and I"ll use the relations $2\sin\frac {B + C}{2}\cos\frac {B - C}{2} = \sin B + \sin C$ and $\cos^2\frac A2 = \frac {s(s-a)}{bc}$ $\implies$

$(\sin B + \sin C)^2\le \frac {(b + c)^2}{b^2 + c^2}\cdot\frac {2s(s - a)}{bc}$ . But $\sin B = \frac {b}{2R}$ , $\sin C = \frac {c}{2R}$ $\Longleftrightarrow\ \left\|\ bc\left(b^2 + c^2\right)\le 4R^2\cdot 2s(s - a)\ \right\|$ . Multiply by $r$ and use

the relation $4Rsr = abc$ $\Longleftrightarrow\ \left\|\ rbc\left(b^2 + c^2\right)\le R\cdot abc\cdot 2(s - a)\ \right\|$ . Simplify by $bc > 0$ and $2(s - a) = b + c - a$ $\Longleftrightarrow\ \boxed {\frac rR\le\frac {a(b + c - a)}{b^2 + c^2}\ }$

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

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

Remark. For the last one on right side : $\boxed {\ \frac {2a(s - b)(s - c)}{s\left(b^2 + c^2\right)}\ \le\ \frac rR\ }\Longleftrightarrow$ $\frac {2a(s - b)(s - c)}{s\left(b^2 + c^2\right)}\le\frac {4(s - a)(s - b)(s - c)}{abc}\Longleftrightarrow$

$2a^2bc\le 4s(s - a)(b^2 + c^2)\Longleftrightarrow$ $2a^2bc\le \left[(b + c)^2 - a^2\right](b^2 + c^2)$ $\Longleftrightarrow a^2(b + c)^2\le (b + c)^2(b^2 + c^2)\Longleftrightarrow a^2\le b^2 + c^2$ $\Longleftrightarrow \boxed {\ A\le 90^{\circ}\ }$ .

Proof 6 (Tiger100). Show easily that $\cos (B-C)=\frac {b\cdot\cos B+c\cdot\cos C}{a}$ . Therefore, $\cos (B-C)\le\frac {2bc}{b^2+c^2}$ $\Longleftrightarrow$

$\frac {b^2\cdot 2ac\cdot\cos B+c^2\cdot 2ab\cdot\cos C}{2a^2bc}\le \frac {2bc}{b^2+c^2}$ $\Longleftrightarrow (b^2+c^2)\left[b^2(a^2+c^2-b^2)+c^2(a^2+b^2-c^2)\right]\le 4a^2b^2c^2$ .

Denote $\left\{\begin{array}{c} a^2=x\\\ b^2=y\\\ c^2=z\end{array}\right\|$ . Thus, $(y+z)\left[x(y+z)-(y-z)^2\right]\le 4xyz\Longleftrightarrow$ $x\left[(y+z)^2-4yz\right]\le (y+z)(y-z)^2$ $\Longleftrightarrow$

$x(y-z)^2\le (y+z)(y-z)^2\Longleftrightarrow y=z\ \vee\ x\le y+z$ $\Longleftrightarrow b=c\ \vee\ a^2\le b^2+c^2\Longleftrightarrow \boxed {\ B=C\ \vee\ A\le 90^{\circ}\ }$ .

Proof 7 (trigonometric - own). $\cos (B-C)\le\frac {2bc}{b^2+c^2}\iff$ $2\cos^2\frac {B-C}{2}\le$ $\frac {(b+c)^2}{b^2+c^2}\iff$ $(\sin B+\sin C)^2\le$ $\frac {(b+c)^2}{b^2+c^2}\cdot (1+\cos A)\iff$

$b^2+c^2\le 4R^2(1+\cos A)\iff$ $\sin^2B+\sin^2C\le$ $1+\cos A\iff$ $\cos 2B+\cos 2C+2\cos A\ge 0\iff$ $2\cos A [1-\cos (B-C)]\ge 0$ .

Proof 8 (synthetic - own). Denote the second intersection $S$ of the $A$ -bisector with the circumcircle $w$ of $\triangle ABC$ and $SB=SC=l$ , $AS=L$ .

Suppose w.l.o.g. $b\ne c$ . Apply the Ptolemy's theorem to $ABSC$ and obtain that $l(b+c)=aL\ (*)$ . Denote the reflection $A'$ of $A$ w.r.t. the diameter $[NS]$ of $w$ .

Thus, $NS=\frac {|b^2-c^2|}{a}$ and $NS^2=2L^2\left[1-\cos (B-C)\right]$ . Therefore, $\cos (B-C)\le\frac {2bc}{b^2+c^2}\iff$ $\frac {\left(b^2-c^2\right)^2}{2(aL)^2}\ge$ $1-\frac {2bc}{b^2+c^3}\iff$

$(b-c)^2(b+c)^2\left(b^2+c^2\right)\ge$ $2a^2L^2(b-c)^2\stackrel{(*)}{\iff}$ $(b+c)^2\left(b^2+c^2\right)\ge$ $2l^2(a+b)^2\iff$ $b^2+c^2\ge$ $2l^2$ ,

what is truly because $A\le 90^{\circ}\le$ $m\left(\widehat{BSC}\right)\iff$ $2l^2\le a^2\le b^2+c^2\implies$ $b^2+c^2\ge 2l^2$ .

Remark. Prove that if in $\triangle ABC$ we have $A\le 90^{\circ}$ and $a\not\in (b,c)\cup (c,b)$ , then $\boxed{\frac bc+\frac cb\le\frac {a^2}{2rr_a}\le\frac Rr}$ (see PP5).

Indeed, $\frac bc+\frac cb\le\frac {a^2}{2rr_a}\iff$ $2\cdot\frac {S^2}{s(s-a)}\cdot\left(b^2+c^2\right)\le a^2bc\iff$ $4(s-b)(s-c)\left(b^2+c^2\right)\le 2a^2bc\iff$

$\left[a^2-(b-c)^2\right]\left(b^2+c^2\right)\le 2a^2bc\iff$ $a^2(b-c)^2\le (b-c)^2\left(b^2+c^2\right)\iff$ $(b-c)^2\left(b^2+c^2-a^2\right)\ge 0\iff$ $\boxed{(b-c)^2\cdot\cos A\ge 0}$ .

$\frac {a^2}{2rr_a}\le\frac Rr\iff$ $a^2\le 2Rr_a\iff$ $2a^2(s-a)\le 4RS\iff$ $2a(s-a)\le bc\iff$ $a(b+c-a)\le bc\iff$ $\boxed{(a-b)(a-c)\ge 0}$ .

$\blacktriangleright$ PP4. $\triangle ABC\ :\ 1\ge \boxed{\cos\frac {B-C}{2}\ge\sqrt {\frac {2r}{R}}}$ .

Proof 1 (trigonometric). $\cos\frac {B-C}{2}\ge \sqrt{\frac {2r}{R}}\iff$ $2\sin\frac {B+C}{2}\cos\frac {B-C}{2}\ge 2\cos\frac A2\cdot\sqrt {\frac {2r}{R}}\iff$ $\sin B+\sin C\ge2\cdot\sqrt {\frac {2r}{R}\cdot\frac {s(s-a)}{bc}}\iff$

$b+c\ge 4R\cdot\sqrt {\frac {2r}{R}\cdot\frac {as(s-a)}{abc}}\iff$ $b+c\ge \sqrt {16R^2\cdot \frac {2r}{R}\cdot\frac {as(s-a)}{4Rsr}}\iff$ $(b+c)^2\ge 4a(b+c-a)\iff$

$(b+c)^2-4a(b+c)+4a^2\ge 0\iff$ $(b+c-2a)^2\ge 0$ , what is truly.

Proof 2 (synthetic). Denote the second intersection $S$ of the (interior) $A$ -bisector with the circumcircle of $\triangle ABC$ . Thus, $2R\cos\frac {B-C}{2}=$

$AS=IA+IS\ge 2\sqrt {IA\cdot IS}=$ $2\sqrt {2Rr}\implies$ $\cos\frac {B-C}{2}\ge\sqrt {\frac {2r}{R}}$ . In conclusion, $\boxed{\sqrt {\frac {2r}{R}}\le \cos\frac {B-C}{2}\stackrel{\left(A\le 90^{\circ}\right)}{\ \le\ }\frac {b+c}{\sqrt {2(b^2+c^2)}}}\le 1\ (*)$ .

$\blacktriangleright$ PP5. $\triangle ABC\ :\ 2\le \boxed{\frac cb+\frac bc\le\frac Rr}$ (Viorel Bandila).

Proof 1. $l^2_a=$ $\frac {4bcs(s-a)}{(b+c)^2}\le s(s-a)$ a.s.o. $\implies$ $h^2_a\le s(s-a)$ a.s.o. $\implies$ $h^2_b+h^2_c\le$ $s(s-b)+s(s-c)=as$ $\implies$

$4S^2\left(\frac {1}{b^2}+\frac {1}{c^2}\right)\le as\implies$ $4S^2\left(\frac cb+\frac bc\right)\le sabc=4sRS\implies$ $sr\left(\frac cb+\frac bc\right)\le sR\implies$ $\frac cb+\frac bc\le\frac Rr$ .

Proof 2 (algebraic). If $\{x,y.z\}\subset (0,\infty )$ , then $\left\{\begin{array}{ccc} \frac {1}{x+y}\le\frac {x+y}{4xy} & \iff & \frac {x+z}{x+y}\le\frac {(x+y)(x+z)}{4xy}\cdot\frac zz\\\\ \frac {1}{x+z}\le\frac {x+z}{4xz} & \iff & \frac {x+y}{x+z}\le\frac {(x+z)(x+y)}{4xz}\cdot\frac yy\end{array}\right\|\ \bigoplus\ \implies$

$\frac {x+z}{x+y}+\frac {x+y}{x+z}\le$ $\frac {(x+y)(x+z)(y+z)}{4xyz}\stackrel{(\mathrm{Ravi})}{\implies}\frac bc+\frac cb\le \frac {abc}{4(s-a)(s-b)(s-c)}=\frac {4Rsr}{4sr^2}\implies$ $\frac bc+\frac cb\le\frac Rr$ .

An easy extension (own). $\boxed{\frac cb+\frac bc\le\frac cb\cdot\frac {2\left(a^2+c^2\right)}{(a+c)^2}+\frac bc\cdot\frac {2\left(a^2+b^2\right)}{(a+b)^2}\stackrel{\mathrm{(ABC\ -\ acute)}}{\le} \frac {c}{b\cdot\cos^2\frac {A-C}{2}}+\frac {b}{c\cdot\cos^2\frac {A-B}{2}}\le\frac Rr}$ .

Proof. $l^2_a=$ $\frac {4bcs(s-a)}{(b+c)^2}\le s(s-a)$ a.s.o. $\implies$ $l^2_b+l^2_c\le$ $s(s-b)+s(s-c)=as$ $\implies$

$4S^2\left[\frac {1}{b^2\cdot\sin^2\left(A+\frac B2\right)}+\frac {1}{c^2\cdot\sin^2\left(B+\frac C2\right)}\right]\le as\implies$ $4S^2\left(\frac {c}{b\cdot\cos^2\frac {A-C}{2}}+\frac {b}{c\cdot\cos^2\frac {A-B}{2}}\right)\le sabc=4sRS\implies$

$\boxed{\frac {c}{b\cdot\cos^2\frac {A-C}{2}}+\frac {b}{c\cdot\cos^2\frac {A-B}{2}}\le\frac Rr}$ . Using the chain of inegalities from $(*)$ obtain easily the required chain,

$\blacktriangleright$ PP6. $\triangle ABC\ ,\ A=90^{\circ}\ :\ \boxed{h_a+\max\{b,c\}\le\frac {3a\sqrt 3}{4}}$ (Virgil Nicula).

Proof 1 (synthetic). Denote the second intersection $D$ of the circumcircle $C(O,R)$ with the $A$ -altitude of $\triangle ABC$ .

Apply in $\triangle ABC$ the remarkable inequality $2s\le 3R\sqrt 3$ , where $s=h_a+b$ and $a=2R\ \ :\ \ h_a+b\le \frac {3a\sqrt 3}{4}$ .

Proof 2 (by extremum). Suppose w.l.o.g. that $c\le b$ and $a$ is constant. Thus $b^2+c^2=a^2$ (constant) and $\left(h_a+b\right)\ -\ \max\ \iff$

$\left(\frac {bc}{a}+b\right)\ -\ \max\ \iff$ $b(a+c)\ -\ \max\ \iff$ $b^2(a+c)^2\ -\ \max\ \iff$ $(a-c)(a+c)^3\ -\ \max\ \iff$

$(a-c)=\frac {a+c}{3}=\frac a2$ because $(a-c)+3\cdot\frac {a+c}{3}=2a$ (constant). Thus, $c=\frac a2$ , $b=\frac {a\sqrt 3}{2}$ , $h_a=\frac {a\sqrt 3}{4}$ and $h_a+b=\frac {3a\sqrt 3}{4}$ .

$\blacktriangleright$ PP7. Prove that $\boxed{\frac{m_{a}}{h_{a}}\ge\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)}$ , where $m_{a}$ , $h_{a}$ are the lengths of the $A$ -median and the $A$ -altitude respectively of the given $\triangle ABC$ .

Remark.This inequality is well-known and is equivalently with $\boxed{\frac {b^{2}+c^{2}}{4R}\le m_a}\stackrel{(\mathrm{acute})}{\ \le\ }\frac {a(s-a)}{2r}$ .

Proof 1. Let diameters $[AA']$ , $[BB']$ in circumcircle $C(O,R)$ of $\triangle ABC$ . Apply theorem of median in $\triangle AMA'$ to $A$ -median, i.e. $B'C^2=4\cdot MO^{2}=$

$2\left(m^{2}_a+A'M^{2}\right)-4R^{2}$ $\Longleftrightarrow$ $4R^{2}-a^{2}=2\cdot\left(m_a^{2}+A'M^{2}\right)-4R^{2}$ $\Longleftrightarrow$ $\boxed{2\left(AM^{2}+A'M^{2}\right)=8R^{2}-a^{2}}\ (*)$ . But $AM+A'M\ge 2R$

with equality iff $b=c$ or $A=90^{\circ}\iff AM+A'M\ge 2R$ $\Longleftrightarrow$ $2\cdot \left(AM^{2}+A'M^{2}\right)+4\cdot AM\cdot A'M\ge 8R^{2}$ $\stackrel{(*)}{\Longleftrightarrow}$ $\boxed{4\cdot AM\cdot A'M\ge a^{2}}\ (1)$ .

Observe that $\left\{\begin{array}{c}A'B^{2}=4R^{2}-c^{2}\\\\ A'C^{2}=4R^{2}-b^{2}\end{array}\right\|$ and $bc=2Rh_{a}$ . Apply theorem of median to $A$ -median in $\triangle ABC$ and to $A'$ -median in $\triangle A'BC$ :

$\left\{\begin{array}{c}4\cdot AM^{2}=2\left(b^{2}+c^{2}\right)-a^{2}\\\\ 4\cdot A'M^{2}=2\left(8R^{2}-b^{2}-c^{2}\right)-a^{2}\end{array}\right\|\ \stackrel{(1)}{\implies}$ $\left[2\left(b^{2}+c^{2}\right)-a^{2}\right]\cdot\left[2\left(8R^{2}-b^{2}-c^{2}\right)-a^{2}\right]\ge a^{4}$ $\Longleftrightarrow$

$\left(b^{2}+c^{2}\right)\left(8R^{2}-b^{2}-c^{2}\right)\ge 4a^{2}R^{2}$ $\Longleftrightarrow$ $4R^{2}\left[2\left(b^{2}+c^{2}\right)-a^{2}\right]\ge \left(b^{2}+c^{2}\right)^{2}$ $\Longleftrightarrow$ $\boxed{\ 4Rm_{a}\ge b^{2}+c^{2}\ }$ $\Longleftrightarrow$

$2\cdot 2Rh_{a}\cdot m_{a}\ge h_{a}\left(b^{2}+c^{2}\right)$ $\Longleftrightarrow$ $2\cdot bc\cdot m_{a}\ge h_{a}\left(b^{2}+c^{2}\right)$ $\Longleftrightarrow$ $\boxed{\ \frac{m_{a}}{h_{a}}\ge \frac{b^{2}+c^{2}}{2bc}\ }$ , with equality if and only if $b=c$ or $A=90^{\circ}$ .

Proof 2. Suppose w.l.o.g. $b\ne c$ . Denote $\phi=m\left(\widehat{AMB}\right)$ and $D\in BC$ for which $AD\perp BC$ . Thus, $\left|b^2-c^2\right|=\left|DB^2-DC^2\right|=$ $a|DB-DC|=$

$2a\cdot MD\implies$ $\boxed{MD=\frac {\left|b^2-c^2\right|}{2a}}$ and $\cot\phi =\frac {DM}{DA}=$ $\frac {\left|b^2-c^2\right|}{2ah_a}\implies$ $\boxed{\cot\phi =\frac {\left|b^2-c^2\right|}{4S}}$ . Thus, $bc\ge 2S\iff$ $\frac {\left|b^2-c^2\right|}{2S}\ge \frac {\left|b^2-c^2\right|}{bc}\iff$

$2\cot\phi\ge\left|\frac bc-\frac cb\right|\iff$ $4\cot^2\phi\ge\left(\frac bc+\frac cb\right)^2-4\iff$ $4\left(1+\cot^2\phi\right)\ge \left(\frac bc+\frac cb\right)^2\iff$ $\frac{2}{\sin\phi}\ge \frac bc+\frac cb\iff$ $\frac {m_a}{h_a}\ge\frac 12\cdot\left(\frac bc+\frac cb\right)$ .

Remark. The proposed inequality can write $\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\le \frac{m_{a}}{h_{a}}$ . It is well-known Bandila's inequality $\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\le \frac{R}{2r}$ , with equality iff $b=c$ .

From $m_a\le R+OM$ (in an acute triangle) obtain that $m_a\le R(1+\cos A)\iff$ $m_a\le2R\cos^2\frac A2\iff$ $2rm_a\le4Rr\cos^2\frac A2=$

$4R(s-a)\tan\frac A2\cos^2\frac A2=$ $4R(s-a)\sin\frac A2\cos\frac A2=$ $2R(s-a)\sin A=a(s-a)\implies$ $\boxed{m_a\le\frac {a(s-a)}{2r}}$ .

Observe that $2r\cdot \sum m_a\le\sum a(s-a)=2r(4R+r)\implies$ $\sum m_a\le 4R+r\implies$ $\boxed{m_a+m_b+m_c\le r_a+r_b+r_c}$ and

$2R\cdot\sum m_a\ge \sum \frac {b^2+c^2}{2}=\sum a^2\implies$ $\boxed{m_a+m_b+m_c\ge\frac {a^2+b^2+c^2}{2R}}$ . From $m_a\le R(1+\cos A)$ a.s.o. obtain that

$\sum am_a\le\sum aR(1+\cos A)=$ $2Rs+R\cdot\sum a\cdot\cos A=2Rs+2R^2\cdot\sum\sin A\cos A=$ $2Rs+R^2\cdot\sum\sin 2A=$ $2Rs+$

$4R^2\cdot \prod\sin A=$ $2Rs+2S=$ $2Rs+2sr$ $\implies$ $6sr\le\boxed{am_a+bm_b+cm_c\le 2s(R+r)}$ . From $4m_a^2=2\left(b^2+c^2\right)-a^2\ge$

$(b+c)^2-a^2=$ $4s(s-a)\implies$ $\boxed{m_a\ge s(s-a)}\ge l_a\ge h_a$ . In conclusion, $\boxed {m_a^2+m_b^2+m_c^2\ge s^2}$ $\ge\sum l_a^2$ .

$\boxed{\blacktriangleleft\ *\ \blacktriangleright}$ Research theme. Compare the ratios $\frac{m_{a}}{h_{a}}$ and $\frac{R}{2r}$ . See here

$\blacktriangleright$ PP8. The $A$ -bisector of $\triangle ABC$ intersects the circumcircle of $\triangle ABC$ second time at $A'$ . Prove that $A'A^2-A'B^2=bc$ .

Proof 1 (synthetic). Denote incenter $I$ and exincenter $I_a$ of $\triangle ABC$ . Prove easily that $A'B=A'C=A'I=A'I_a$ , i.e. the quadrilateral

$IBI_aC$ is inscribed in a circle $w$ with the center $A'$ and the radius $A'B$ . Observe that $\triangle AIC\sim\triangle ABI_a$ . In conclusion, $\frac {AC}{AI_a}=\frac {AI}{AB}$ ,

i.e. $\boxed{AI\cdot AI_a=bc}\ (*)$ . The power of $A$ w.r.t. $w$ is $p_w(A)=AI\cdot AI_a=A'A^2-A'B^2\stackrel{(*)}{\iff}$ $A'A^2-A'B^2=bc$ .

Proof 2 (synthetic). Let $D\in AI\cap BC$ . Prove easily that $\triangle ABD\sim \triangle AA'C$ . So $\frac{AB}{AA'}=\frac{AD}{AC}\Longleftrightarrow \boxed{AD\cdot AA'=bc} \ (1)$ .

Since the line $A'B$ is tangent to the circumcircle of $\triangle ABD$ obtain that $A'D\cdot A'A=A'B^2\ (2)$ . Adding these two relations

$(1)$ and $(2)$ side by side obtain that $A'A\cdot (AD+A'D)=A'B^2+bc\iff$ $A'A^2-A'B^2=bc$ .

Proof 3 (metric). $A'A^2-A'B^2=$ $A'A^2-A'I^2=$ $IA(A'A+A'I)=$ $AI\cdot AI_a$ $\implies$ $\boxed{A'A^2-A'B^2=AI\cdot AI_a}$ .

$\sqrt{\frac {s(s-a)}{bc}}=\cos\frac A2=$ $\frac {s-a}{AI}=\frac {s}{AI_a}=$ $\sqrt{\frac {s(s-a)}{AI\cdot AI_a}}\implies$ $\boxed{AI\cdot AI_a=bc}$ . In conclusion, $A'A^2-A'B^2=bc$ .

Proof 4 (metric). $A'A^2-A'B^2=$ $A'A^2-A'I^2=$ $(A'A-A'I)[(A'I+IA)+A'I]=$ $AI\cdot (AI+2\cdot A'I)=$ $AI^2+$

$2\cdot IA\cdot IA'=$ $\frac {bc(s-a)}{s}+4Rr=$ $bc-\frac {abc}{s}+4Rr=$ $bc-4Rr+4Rr=bc$ . In conclusion, $A'A^2-A'B^2=bc$ .

Proof 5 (trigonometric). $A'A^2-A'B^2=4R^2\cdot\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$ $4R^2\sin\left(B+\frac A2+\frac A2\right)\sin\left(B+\frac A2-\frac A2\right)=$

$4R^2\sin (B+A)\sin B=$ $4R^2\sin C\sin B=$ $2R\sin B\cdot 2R\sin C=AC\cdot AB\implies$ $A'A^2-A'B^2=bc$ .

Lemma. Let $\{A,C,B,D\}\subset d$ be four points on the line $d$ in this order. Then $\frac {\overline{CA}}{\overline{CB}}=-\frac {\overline{DA}}{\overline{DB}}$ - "harmonical

division" on $d$ $\iff$ $\boxed{MA^2-MC^2=\overline{AC}\cdot \overline{AD}=\overline{AB}\cdot \overline{AM}}$ , where $M$ is the midpoint of $[CD]$ .

Proof. Define by $X(x)$ , the point $X$ with the affix $x$ . Then $\overline{XY}=y-x$ and $2m=d-c$ . Prove easily that $\frac {\overline{CA}}{\overline{CB}}=-\frac {\overline{DA}}{\overline{DB}}\iff$

$(a+b)(c+d)=2(ab+cd)\iff$ $(c-a)(d-a)=(b-a)\left(\frac {c+d}{2}-a\right)\iff$ $\overline{AC}\cdot \overline{AD}=\overline{AB}\cdot \overline{AM}$ .

Proof 6. Observe that $\frac {DI}{DI_a}=\frac {r}{r_a}=$ $\frac {s-a}{s}=\frac {AI}{AI_a}$ , what means that the points $\{A,I,D,I_a\}$ form an harmonical division

on the $A$ -bisector of $\triangle ABC$ , where $A'$ is the midpoint of $[II_a]$ . Using upper lemma obtain that $AI\cdot AI_a=AD\cdot AA'=bc$ .

$\blacktriangleright$ PP9. Let $ABCD$ be a cyclical convex quadrilateral. Denote the midpoints $E$ , $F$ , $G$ , $H$ of corresponding arcs for the chords

$[AB]$ , $[BC]$ , $[CD]$ , $[DA]$ respectively. Suppose that $EG\cdot FH=AC\cdot BD$ . Prove hat $EG\cap FH\cap AC\cap BD\ne\emptyset$ .

Proof. Denote the length $r$ of the circumradius. Observe that

$\left\{\begin{array}{c} EF\cdot GH=2r\sin \frac D2\cdot 2r\sin \frac B2=4r^2\sin\frac B2\cos\frac B2=2r^2\cdot \sin B=r\cdot AC\\\\ EH\cdot FG=2r\sin \frac C2\cdot 2r\sin \frac A2=4r^2\sin\frac A2\cos\frac A2=2r^2\cdot \sin A=r\cdot BD\end{array}\right\|$ . Apply the Ptolemy's theorem to $EFGH\ :$

$r(AC+BD)=EF\cdot GH+EH\cdot FG=$ $EG\cdot FH=AC\cdot BD$ $\implies$ $AC\cdot BD=r\cdot (AC+BD)\ge$ $2r\sqrt {AC\cdot BD}$

$\implies$ $4r^2\ge AC\cdot BD\ge 4r^2\implies$ $AC\cdot BD=4r^2\implies$ $AC=BD=2r\implies$ $ABCD$ is rectangle $\implies$ $EG\cap FH\cap AC\cap BD\ne\emptyset$ .

PP10. Let $ABC$ be an acute triangle with incenter $I$ , circumcircle $w=C(O,R)$ and

orthocenter $H$ . Let $\{A,A'\}=AI\cap w$ . Prove that $A=60^{\circ}\implies$ $AA'=HB+HC$ .

Proof 1 (trigonometric). Observe that $AA'=2R\sin\left(B+\frac A2\right)=2R\cos\frac {B-C}{2}$ and $HB+HC=2R(\cos B+\cos C)=$

$4R\cos\frac {B+C}{2}\cos \frac {B-C}{2}=2R\cos\frac {B-C}{2}$ because $B+C=120^{\circ}$ . In conclusion, $HB+HC=AA'$ .

Proof 2 (synthetic). Denote the second intersection $E$ of $BH$ with the circle $w$ . Observe that the triangle $HEC$ is equilateral and $ABA^{\prime}E$

is a trapezoid with $AB\parallel A^{\prime}E$ , from where obtain that $AA^{\prime}=BE=BH+HE=BH+HC$ , i.e $AA^{\prime}=HB+HC$ .

Remark. Denote the second intersection $F$ of $CH$ with the circle $w$ . Prove easily that : $BF\parallel AA'\parallel CE$ ; $A'E\parallel AB$ and $A'E=b$ ;

$A'F\parallel AC$ and $A'F=c$ ; $AH=AF=A'C=A'B=AE$ ; $AA'=BE=CF=HB+HC$ ; $\boxed{HA=HA'}$ .

An equivalent enunciation. Let $ABC$ be a triangle with $A=120^\circ$ , the circumcenter $O$ and the orthocenter $H$ . Prove that $OH=AB+AC$ .

Proof (metric). $a=R\sqrt 3\ ,\ a^2=b^2+c^2+bc\ ,\ GO^2=R^2-\frac {a^2+b^2+c^2}{9}$ and $HO=3\cdot GO\implies$ $GO^2=\frac {a^2}{3}-\frac {\sum a^2}{9}=$

$\frac {2a^2-b^2-c^2}{9}=$ $\frac {2(b^2+c^2+bc)-b^2-c^2}{9}=$ $\frac {b^2+c^2+2bc}{9}=$ $\frac {(b+c)^2}{9}\implies$ $GO= \frac {b+c}{3}$ $\implies$ $HO=b+c$ .

PP11 (Selection Team of BOSNIA-HERTEGOVINA, 2011). Let $ABC$ be a triangle for which $2a=b+c$ . Denote

its incenter $I$ and the midpoints $M$ , $N$ of the sides $[AB]$ , $[AC]$ respectively. Prove that $AMIN$ is a cyclical quadrilateral.

Proof 1. Denote the circumcircle $w=C(O,R)$ for $\triangle ABC\ ,\ \{A,S\}=AI\cap w$ and the diameter $[NS]$ of $w$ . Hence $IA=IS\iff$ $IA^2=IA\cdot IS\iff$

$IA^2=2Rr\iff$ $bc-4Rr=2Rr\iff$ $bc=6Rr\iff$ $2sbc=3\cdot 4Rsr=3abc\iff$ $2s=3a\iff b+c=2a$ . In conclusion,

$2a=b+c\iff$ $I$ is the midpoint of $[AS]\iff IM\parallel SB\ \ \wedge\ \ IN\parallel SC\iff$ $\widehat {MIN}\equiv\widehat{BSC}\iff$ $AMIN$ is a cyclical quadrilateral.

I used the notation $2s=a+b+c$ , an well-known relation $AI^2=\frac {bc(s-a)}{s}=bc-\frac {abc}{s}=bc-4Rr$ and the power of $I$ w.r.t. $w\ : p_w(I)=-2Rr$ .

Remark. $IG\parallel BC\iff h_a=3r\iff ah_a=3ar \iff$ $2sr=3ar\iff 2s=3a\iff b+c=2a$ . In the triangle $ABC$ avem $2a=b+c$ .

Otherwise. Is well-known $AD\cdot AS=bc$ , where $D\in AI\cap BC$ . The incenter $I$ is the midpoint of $AS]\iff$ $bc=AD\cdot AS=2\cdot AD\cdot AI=$

$AD^2\cdot \frac {b+c}{s}\iff$ $bc=\frac {b+c}{s}\cdot \frac {4bcs(s-a)}{(b+c)^2}\iff$ $b+c=4(s-a)\iff b+c=2(b+c-a)\iff b+c=2a$ .

PP12 (France, 1999). Let $\triangle ABC$ , $H$ - its orthocenter, $O$ - its circumcenter, $R$ - its circumradius. Let $A_1$ be the reflection of $A$ across $BC$ .

Let $B_1$ be the reflection of $B$ across $CA$ . Let $C_1$ be the reflection of $C$ across $AB$ . Prove that the points $A_1\in B_1C_1\ \iff\ OH=2R$ .

Proof. Let $X,Y,Z$ be the projections of the nine-point center $\mathcal E$ on $BC,CA,AB$ respectively. If $M,N,P$ are the midpoints of $BC,CA,AB$ respectively and $D,E,F$

are the feet of the perpendiculars from $A,B,C$ on the opposite sides, then $X,Y,Z$ are the midpoints of $MD,NE,PF$ respectively. Use vectors to show that

$XY\parallel A_1B_1$ , $YZ\parallel B_1C_1$ and $ZX\parallel C_1A_1$ . This means that $A_1,B_1,C_1$ are collinear $\Longleftrightarrow$ $X,Y,Z$ are collinear $\Longleftrightarrow$ the projections of $\mathcal E$ on the sides of

$\triangle\ ABC$ are collinear $\Longleftrightarrow \mathcal E$ lies on the circumcircle of $ABC$ (using reciprocally the Simson's line) $\Longleftrightarrow$ $\frac {OH}{2} = R$ , i.e. $OH = 2R$ . I"ll show now only what

is missing from the Grobber's proof, i.e. $A_1B_1\parallel XY$ . Indeed, since the nine-point center is the middlepoint of the segment $[OH]$ , obtain

$\left\{\begin{array}{ccccc} 2\cdot\overline {\mathcal EX} & = & \overline {HD} + \overline {OM} & = & \overline {HD} + \frac 12\cdot\overline {AH} \\ \\ 2\cdot \overline {\mathcal EY} & = & \overline {HE} + \overline {ON} & = & \overline {HE} + \frac 12\cdot\overline {BH}\end{array}\right\|$ $\implies$ $\frac {\overline {\mathcal EX}}{\overline {\mathcal EY}} =$ $\frac {\overline {HD} + \left(\overline {AH} + \overline {HD}\right)}{\overline {HE} + \left(\overline {BH} + \overline {HE}\right)} =$ $\frac {\overline {HD} + \overline {AD}}{\overline {HE} + \overline {BE}} =$ $\frac {\overline {HD} + \overline {DA_1}}{\overline {HE} + \overline {EB_1}} =$ $\frac {\overline {HA_1}}{\overline {HB_1}}$ $\implies$

$\frac {\overline {\mathcal EX}}{\overline {\mathcal EY}} =$ $\frac {\overline {HA_1}}{\overline {HB_1}}$ , i.e. $\triangle\ X\mathcal EY\sim\triangle\ A_1HB_1$ . In conclusion, $A_1B_1\parallel XY\ .$ I like very much the Grobber's proof.

PP13. Stronger than Nesbitt's inequality: $\boxed{\ \frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\ge\frac 52-\frac {2\cdot\sum bc}{\sum a^2+\sum bc}\ge \frac 52-\sqrt{\frac {\sum bc}{\sum a^2}}\ge \frac 32\ }\ .$

$\blacktriangleright\ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}$ $+\frac{2(ab+bc+ac)}{a^2+b^2+c^2+ab+bc+ac}=$ $\sum\frac{ a^2}{a(b+c)}+\frac{2\sum ab}{\sum a^2+\sum ab}\ge$

$\frac{(\sum a)^2}{2\sum ab}+\frac{2\sum ab}{\sum a^2+\sum ab}=$ $\frac{1}{2}+\frac{\sum a^2+\sum ab}{2\sum ab}+\frac{2\sum ab}{\sum a^2+\sum ab}\ge$ $\frac{1}{2}+2=\frac{5}{2}\ .$

$\blacktriangleright\ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}\ge$ $\frac{(\sum a)^2}{2\sum ab}+\sqrt{\frac{\sum ab}{\sum a^2}}=$ $1+\frac{\sum a^2}{2\sum ab}+\sqrt{\frac{\sum ab}{\sum a^2}}=$

$1+\frac{1}{2}\left(\frac{\sum a^2}{\sum ab}+\sqrt{\frac{\sum ab}{\sum a^2}}+\sqrt{\frac{\sum ab}{\sum a^2}}\right)\ge$ $1+\frac{3}{2}=\frac{5}{2}\ .$ [/color]

PP14. Let $ABCD$ be a rhombus. Let $P\in (BC)$ and $Q\in (CD)$ such that $BP = CQ$ . Prove that the centroid of $\triangle APQ$ lies on $(BD)$ .

Proof 0 (synthetic - Andrei Theodor, very nice !). Denote $X\in AP\cap BD$ and $Y\in AQ\cap BD$ . Observe that $\frac {XP}{XA}+\frac {YQ}{YA}=$

$\frac {BP}{AD}+\frac {DQ}{AB}=$ $\frac {BP}{BC}+\frac {PC}{BC}=1$ . From an well-known property obtain that the centroid $G$ of $\triangle ABC$ belongs to $XY$ , i.e. $G\in BD$ .

Proof 1 (synthetic). Denote $R\in (AD)$ so that $AR=BP=CQ$ , $S\in QR\cap BD$ , the midpoint $T$ of $[AQ]$ and $G\in BD\cap PT$ . Observe that $ACQR$

is an isosceles trapezoid, $TS$ is the $Q$ -middle line in $\triangle AQR$ and $TS\parallel BP$ with $\frac {GP}{GT}=\frac {BP}{TS}=\frac {AR}{TS}=2$ , i.e. $G$ is the centroid of the triangle $APQ$ .

Proof 2 (with vectors). Denote $M\in (BD)$ so that $CPMQ$ is a parallelogram. Observe that $BP=PM=CQ$ . Thus,

$\overrightarrow{AP} -\overrightarrow{AM}=$ $\overrightarrow{MP}=$ $\overrightarrow{QC}=\overrightarrow{AC}-\overrightarrow{AQ}$ , i.e. the triangles $APQ$ and $AMC$ have a common $A$ -median $AS$ , where

$S\in PQ\cap CM$ . Hence these triangles have and a common centroid $G$ , where $G\in AS\cap MD$ , i.e. $G\in BD$ .

Proof 3 (analytic). Suppose w.l.o.g. $AB=1\ ,\ m\left(\widehat{BAD}\right)=\phi <90^{\circ}$ and $A(0,0)$ , $B(\cos\phi ,\sin\phi )$ , $C(1+\cos\phi , \sin\phi)$

and $D(1,0)$ . For $BP=CQ=r<1$ obtain easily that $P(1-r+\cos\phi , \sin\phi )$ and $Q(1+r\cos\phi , r\sin\phi )$ .Therefore,

the centroid $G_{APQ}\left(\frac {2-r+(1+r)\cos\phi}{3},\frac {(1+r)\sin\phi}{3}\right)\in (BD)$ $\iff$ $\left|\begin{array}{ccc} 1 & 0 & 1\\\\ \cos\phi & \sin\phi & 1\\\\ 2-r+(1+r)\cos\phi & (1+r)\sin\phi & 3\end{array}\right|=0$

$\iff \left|\begin{array}{ccc} 1 & 0 & 1\\\\ \cos\phi & 1 & 1\\\\ 2-r+(1+r)\cos\phi & (1+r) & 3\end{array}\right|=0$ $\iff \left|\begin{array}{ccc} 1 & 0 & 1\\\\ 0 & 1 & 1\\\\ 2-r & 1+r & 3\end{array}\right|+\cos\phi\cdot \left|\begin{array}{ccc} 0 & 0 & 1\\\\ 1 & 1 & 1\\\\ 1+r & 1+r & 3\end{array}\right|=0$ ,

what is truly because we have in the first determinant $C_3=C_1+C_2$ and in the second determinant $C_1=C_2$ .

Otherwise, prove easily that $G\in BD$ $\iff$ $y_G=\frac {\sin\phi}{\cos\phi -1}\cdot (x_G-1)$ , i.e. $(1+r)\sin\phi=\frac {\sin\phi}{\cos\phi -1}\cdot (\cos\phi -1)(1+r)$ .

PP15 (the extension of RMO2, India). Let $\triangle ABC$ with circumcircle $w$ and incentre $I$ . Let $[AA']$ be a diameter

of $w$ . Denote $D\in BC$ for which $ID\perp BC$ and $\{A,S\}=AI\cap w$ . Show that $A'I$ and $SD$ intersect on $w$ .

Proof. Let $L\in SD\cap A'I$ . Thus $\left\{\begin{array}{c} AA'=2R\ ;\ ID=r\\\\ IA\cdot IS\ =\ 2Rr\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c} \frac {ID}{AI}=\frac {IS}{AA'}\\\\ \widehat{SID}\equiv\widehat{A'AI}\end{array}\right\|$ $\stackrel{(s.a.s)}{\implies}$

$\triangle ISD\sim$ $\triangle AA'I\ \implies\ \widehat{ISD}$ $\equiv\widehat{AA'I}\implies AA'SL$ is cyclically $\implies\ L\in w$ .

PP16. Solve for all $x\in\mathbb R$ the equation $\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}$ .

Proof 1. Applying the Cauchy-Schwarz's inequality obtain that $\left(\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}\right)^2=\left(2\sqrt{2}\cdot \frac{1}{\sqrt{x+1}}+\sqrt{x+1}\cdot \frac{\sqrt{x}}{\sqrt{x+1}}\right)^2$ $\le\left[\left(2\sqrt{2}\right)^2+x+1\right]\cdot$

$\left(\frac{1}{x+1}+\frac{x}{x+1}\right)=x+9$ $\Longleftrightarrow$ $\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}\leqslant\sqrt{x+9}$ . In conclusion, $\frac {2\sqrt 2}{\frac {1}{\sqrt {x+1}}}=\frac {\sqrt {x+1}}{\frac {\sqrt x}{\sqrt {x+1}}}\iff$ $2\sqrt {2x}=\sqrt {x+1}\iff$ $\boxed{x=\frac{1}{7}}$ .

Proof 2. $\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\iff$ $\frac{2\sqrt{2}}{\sqrt{x+1}}=\sqrt{x+9}-\sqrt x\iff$ $\frac{2\sqrt{2}}{\sqrt{x+1}}=\frac {9}{\sqrt {x+9}+\sqrt x}\iff$

$2\sqrt 2\left(\sqrt {x+9}+\sqrt x\right)=9\sqrt {x+1}\iff$ $8\cdot \left[2x+9+2\sqrt {x(x+9)}\right]=81(x+1)\iff$ $16\cdot\sqrt {x(x+9)}=65x+9\iff$

$256x(x+9)=(65x+9)^2\iff$ $3969x^2-1134x+81=0\iff$ $49x^2-14x+1=0\iff$ $(7x-1)^2=0\iff$ $\boxed{x=\frac 17}$ .

Proof 3. $\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\iff$ $\frac{2\sqrt{2}}{\sqrt{x+1}}=\sqrt{x+9}-\sqrt x\iff$ $\frac{2\sqrt{2}}{\sqrt{x+1}}=\frac {9}{\sqrt {x+9}+\sqrt x}\iff$

$2\sqrt {2(x+9)}=9\sqrt {x+1}-2\sqrt {2x}\iff$ $8(x+9)=81(x+1)+8x-36\sqrt {2x(x+1)}\iff$ $36\sqrt {2x(x+1)}=81x+9\iff$

$4\sqrt {2x(x+1)}=9x+1\iff$ $32x(x+1)^2=81x^2+18x+1\iff$ $49x^2-14x+1=0\iff$ $(7x-1)^2=0\iff$ $\boxed{x=\frac 17}$ .

PP17 (bzprules). Prove that $\{a,b,c,d\}\subset\mathbb R^*_+\ \implies\ a^4b+b^4c+c^4d+d^4a\ge a^2bcd+ab^2cd+abc^2d+abcd^2$ .

Proof. Let's have some weights $w_1, w_2, w_3, w_4$ , i.e. $w_k\ge 0\ ,\ k\in\overline{1,4}$ and $w_1+w_2+w_3+w_4=1$ . We also want them such that

$w_1a^4b+w_2b^4c+w_3c^4d+w_4d^4a \geq a^2bcd$ and any cyclic permutation of the weights gives another inequality, for example

$w_2a^4b+w_3b^4c+w_4c^4d+w_1d^4a \geq ab^2cd$ a.s.o. If we look at all the 4 cyclic permutations and add those inequalities we get our inequality.

But how to do this? From the weighted-AM-GM inequality, we have that $w_1a^4b+w_2b^4c+w_3c^4d+w_4d^4a \geq$ $\left(a^4b\right)^{w_1}\left(b^4c\right)^{w_2}\left(c^4d\right)^{w_3}\left(d^4a\right)^{w_4}=$

$a^{4w_1+w_4}b^{4w_2+w_1}c^{4w_3+w_2}d^{4w_1+w_3}$ . If we could get the RHS to be equal to $a^2bcd$ , then that'd be awesome. So then it's just solving the system of equations

$\left\{\begin{array}{ccc} 4w_1+w_4 & = & 2\\\ 4w_2+w_1 & = & 1\\\ 4w_3+w_2 & = & 1\\\ 4w_4+w_3 & = & 1\end{array}\right\|$ . Exist an unique solution to this system and we can clearly see that this quad of weights satisfies everything we want. For the inequality,

solving the system of equations yields $\frac {w_1}{23}=\frac {w_2}{7}=\frac {w_3}{11}=$ $\frac {w_1}{10}=\frac {1}{51}\implies$ $\frac{23a^4d+7b^4c+11c^4d+10d^4a}{51} \geq$ $\sqrt[51]{a^{102}b^{51}c^{51}d^{51}} = a^2bcd$ .

This method can be applied in general using the same ideas.

An easy extension. Prove that for any $n\in\mathbb N$ and $\{a,b,c,d\}\subset\mathbb R^*_+$ we have $\boxed{a^{n+3}b+b^{n+3}c+c^{n+3}d+d^{n+3}a\ \ge\ abcd\left(a^n+b^n+c^n+d^n\right)}$ .

Proof. Prove similarly that $:\ \frac {w_1}{n^3+6n^2+11n+5}=\frac {w_2}{2n+5}=$ $\frac {w_3}{n^2+5n+5}=\frac {w_4}{n^2+4n+5}=\frac {1}{(n+2)(n^2+6n+10)}$ .

PP18. Solve the system of the equations $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}+\sqrt{z} =4\\\\ xy+yz-xz=1\\\\ x^{2}+y^{2}+z^{2}=18\end{array}\right\|$ , where $\{x,y,z\}\subset\mathbb R_+$ .

Proof. Obseve that $(x-y+z)^2=\left(x^2+y^2+z^2\right)-2(xy+yz-xz)=18-2=16\implies$ $\boxed{\ x-y+z\in\{\pm 4\}\ }$ . Therefore, $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}+\sqrt{z} =4\\\\ xy+yz-xz=1\\\\ x^{2}+y^{2}+z^{2}=18\end{array}\right\|\iff$

$\left\{\begin{array}{l}\sqrt{x}+\sqrt{z} =4-\sqrt y\\\\ y(x+z)=xz+1\\\\ (x+z)=y\pm 4\end{array}\right\|\iff$ $\left\{\begin{array}{l}\sqrt{x}+\sqrt{z} =4-\sqrt{y}\\\\ x+z=y\pm 4\\\\ xz=y^2\pm 4y-1\end{array}\right\|$ . Denote $\left\{\begin{array}{c} \boxed{S=x+z}=y\pm 4\\\\ \boxed{P=xz}=y^2\pm 4y-1\end{array}\right\|$ . Thus, $0\le y\le 16$ and $\left(\sqrt{x}+\sqrt{z}\right)^2=\left(4-\sqrt{y}\right)^2\iff$

$S+2\sqrt P=16+y-8\sqrt y\iff$ $y\pm 4+2\sqrt {y^2\pm 4y-1}=16+y-8\sqrt y\iff$ $2\sqrt {y^2\pm 4y-1}=16\mp 4-8\sqrt y$ . Denote finally that $\sqrt y=t\ge 0$ , i.e. $y=t^2$ .

In conclusion, appear two cases : $\left\{\begin{array}{c} P=t^4+4t^2-1=\left(6-4t\right)^2\\\\ S=t^2+4\ \ ;\ \ 0<t<\frac 32\end{array}\right\| \ \vee\ \ \left\{\begin{array}{c} P=t^4-4t^2-1=\left(10-4t\right)^2\\\\ S=t^2-4\ \ ;\ \ 0<t<\frac 52\end{array}\right\|$ a.s.o.

PP19. Solve the system of the equations $\left\{\begin{array}{c} \sqrt{x+1}+\sqrt{x+3}+\sqrt {x+5}=\sqrt {y-1}+\sqrt {y-3}+\sqrt {y-5}\\\\ x+y+x^2+y^2=78\end{array}\right\|$ , where $\{x,y,z\}\subset\mathbb R$ .

Proof. $f(x)=\sqrt{x+1}+\sqrt{x+3}+\sqrt {x+5}\ ,\ x\ge -5$ is strict increasing $\implies$ it is injectively. From first equation obtain $f(x)=f(y-6)\iff$ $x=y-6\iff$ $\boxed{y=x+6}$ .

Therefore, the initial system becomes $\left\{\begin{array}{c} y=x+6\\\\ x+y+x^2+y^2=78\end{array}\right\|\implies$ $x^2+7x-18=0\begin{array}{cc} \nearrow & 2\\\\ \searrow & -9\end{array}$ . In conclusion, its solutions are $(x,y)\in \left\{\ (2,8)\ ,\ (-9,-3)\ \right\}$ .

PP20. Prove that $\{a,b,c,d\}\subset\mathbb R$ and $(a+b)(c+d)\le 2(ac+bd)\implies$ the equation $f(x)=(x+a)(x+d)+(x+b)(x+c)=0$ has real roots.

Proof. Prove esily that $(a+b)(c+d)\le 2(ac+bd)\iff (a-b)(c-d)\ge 0$ and $f(-b)f(-d)=-(b-d)^2(a-b)(c-d)\le 0\implies$ the equation $f(x)=0$ has real roots.

This post has been edited 320 times. Last edited by Virgil Nicula, Feb 24, 2019, 9:02 AM

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
this css is sus
by ihatemath123, Aug 14, 2024, 1:53 AM
391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
by GoogleNebula, Apr 14, 2021, 5:25 AM
Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
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

307. Very nice proofs !

by Virgil Nicula, Jul 31, 2011, 11:50 PM

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