67. Remarkable algebraic/geometrical inequalities (1).

by Virgil Nicula, Jul 22, 2010, 11:12 PM

I. Remarkable inequalities: 0

$\bigodot\ \boxed {\ a^2+b^2+c^2\ge 4S\sqrt 3\ }$ (Weitzenbock - see here).

Method I. $16S^2=\sum a^2\left(b^2+c^2-a^2\right)$ and the Cebasev's inequality $\implies 48S^2\ \le \sum a^2\sum\left(b^2+c^2-a^2\right)\ =\ \left(a^2+b^2+c^2\right)^2\ \Longrightarrow\ 4S\sqrt 3\ \le\ a^2+b^2+c^2\ .$

Otherwise. $48S^2=3\cdot\sum\left(a^2+b^2-c^2\right)\left(a^2-b^2+c^2\right)\le$ $\left[\sum\left(b^2+c^2-a^2\right)\right]^2=$ $\left(a^2+b^2+c^2\right)^2\implies$ $4S\sqrt 3\le a^2+b^2+c^2$ .

Remark. $\left\{\begin{array}{c} 4\sin A\cos B\cos C=\sin 2B+\sin 2C-\sin 2A\\\\ 4\sin A\sin B\sin C=\sin 2A+\sin 2B+\sin 2C\\\\ S=2R^2\sin A\sin B\sin C\end{array}\right\|$ $\implies \sum a\cdot\cos B\cos C=$ $2R\cdot \sum \sin A\cos B\cos C=$ $\frac R2\cdot\sum ( \sin 2B+\sin 2C-\sin 2A)=$

$\frac R2\cdot\sum\sin 2A=$ $2R\prod\sin A\implies$ $\boxed{\sum a\cdot\cos B\cos C=\frac SR}\ \ ;\ \ 16S^2=$ $\sum\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)=$

$\sum (2ac\cos B)\cdot(2ab\cos C)\implies$ $4abc\sum a\cos B\cos C=16SR\sum a\cos B\cos C\implies$ $\boxed{\sum a\cos B\cos C=\frac SR}$

Method II. $\sum a^2\ge 4S\sqrt 3$ $\Longleftrightarrow$ $\left(\sum a^2\right)^2\ge 3\cdot\left(2\cdot\sum b^2c^2-\sum a^4\right)$ $\Longleftrightarrow$ $\ \sum a^4+2\cdot\sum b^2c^2\ge 6\cdot\sum b^2c^2-3\cdot\sum a^4\ \Longleftrightarrow\ \sum a^4\ \ge\ \sum b^2c^2$ what is true.

Method III. $OL^2\ =\ R^2\ +\ p_w(L)\ =\ R^2-3\cdot\left(\frac {abc}{a^2+b^2+c^2}\right)^2\ \ge\ 0\ \Longrightarrow$ $R\cdot \sum a^2\ \ge\ abc\sqrt 3$ $\Longrightarrow$ $a^2+b^2+c^2\ \ge \frac {4Rpr\sqrt 3}{R}$ $\Longrightarrow$ $\boxed {\ 4S\sqrt 3\ \le\ a^2+b^2+c^2\ }\ .$

Method IV. I"ll use the inequality $R\ge 2r$ thus: $\left(\sum a^2\right)^2\ge\sum a^2\cdot\sum bc=$ $\ abc\cdot\sum a^2\cdot\sum\frac 1a\ge 3abc(a+b+c)=24Rp^2r\ge 48p^2r^2=48S^2$ $\Longrightarrow$

$a^2+b^2+c^2\ge 4S\sqrt 3\ .$ Using same inequality $R\ge 2r$ can obtain a stronger inequality. Indeed, $\left(4S\sqrt 3\right)^2=24p^2r\cdot 2r\le$ $24p^2r\cdot R=3\cdot 4Rpr\cdot 2p=$

$3abc(a+b+c)\le (ab+bc+ca)^2\Longrightarrow \boxed {\ 4S\sqrt 3\le ab+bc+ca\ }\ .$

Method V. We use the inequality $3r\sqrt 3\le p$ thus : $3r\sqrt 3\le p$ $\Longleftrightarrow$ $3S\sqrt 3\le p^2$ $\Longrightarrow$ $12S\sqrt 3\le (a+b+c)^2\le 3\left(a^2+b^2+c^2\right)$ $\Longrightarrow$ $4S\sqrt 3\le a^2+b^2+c^2$ .

Method VI. We use two inequalities $\left\|\ \begin{array}{c} 2r\ \le\ R\\\\ 2p\ \le\ 3R\sqrt 3\end{array}\ \right\|\ \Longrightarrow\ 4pr\ \le\ 3R^2\sqrt 3\ .$ Prove easily that $4pr\ \le\ 3R^2\sqrt 3\ \Longleftrightarrow$ $4S\sqrt 3\ \le\ 3\cdot\sqrt[3]{a^2b^2c^2}\ .$ But $3\cdot\sqrt[3]{a^2b^2c^2}=3\cdot\sqrt[3]{ab\cdot bc\cdot ca}\ \le\ ab+bc+ca\ .$ Therefore, $\boxed {\ 4S\sqrt 3\ \le\ 3\cdot\sqrt[3] {(abc)^2}\ \le\ ab+bc+ca\ \le\ a^2+b^2+c^2\ }$ (the Weitsenbock's inequality is "weakly" ! ).

Method VII. I"ll use the inequality $p\sqrt 3\le 4R+r\ .$ Indeed, add the inequalities $\ a^2\ge 4(p-b)(p-c)$ etc. $\Longrightarrow$

$\sum a^2\ge 4\sum (p-b)(p-c)=$ $4r(4R+r)\ge 4rp\sqrt 3=4S\sqrt 3$ $\Longrightarrow$ $\boxed {\ 4S\sqrt 3\le 4r(4R+r)\le a^2+b^2+c^2\ }\ .$

Method VIII. $b^2+c^2-a^2=4S\cdot\cot A$ si $\sum\cot B\cot C=1$ $\Longrightarrow$ $\left(\sum\cot A\right)^2\ge 3\sum\cot B\cot C=3\ \Longrightarrow$

$\sum\cot A\ge\sqrt 3\Longrightarrow$ $\sum a^2=\sum\left(b^2+c^2-a^2\right)=4S\cdot\sum\cot A\ge 4S\sqrt 3$ $\Longrightarrow$ $a^2+b^2+c^2\ge 4S\sqrt 3\ .$

Method IX. $a^2+b^2\ge 2ab\ge 2ab\cdot\cos\left(C-60^{\circ}\right)\ \Longrightarrow\ a^2+b^2\ge ab\cdot\cos C+ab\sqrt 3\cdot\sin C\ \Longrightarrow$

$\ 2\left(a^2+b^2\right)-\left(a^2+b^2-c^2\right)\ge 2ab\sqrt 3\cdot\sin C\ \Longrightarrow\ a^2+b^2+c^2\ge 4S\sqrt 3$ because $\left\|\ \begin{array}{c} ab\cdot\sin C=2S\\\\ 2ab\cdot\cos C=a^2+b^2-c^2\end{array}\ \right\|\ .$

Generalization. $\left\|\ \begin{array}{c} m\ >\ 0\\\\ \phi\in\left(0,\pi \right)\end{array}\ \right\|\ \Longrightarrow\ \boxed {\ \left(1-m\cos\phi\right)\cdot a^2+m\left(m-\cos\phi\right)\cdot b^2+m\cos\phi\cdot c^2\ \ge 4m\sin\phi\cdot S\ }$ with the equality iff $m=\frac ab$ and $\phi =C\ .$

Proof. $a^2+m^2b^2\ge 2mab\ge 2mab\cos (C-\phi )=$ $\ 2mab\cos C\cos\phi +2mab\sin C\sin\phi =m\cos\phi\left(a^2+b^2-c^2\right)+4mS\sin\phi\ \Longrightarrow$

$\ \left(1-m\cos\phi\right)\cdot a^2+m\left(m-\cos\phi\right)\cdot b^2+m\cos\phi\cdot c^2\ \ge 4m\sin\phi\cdot S\ .$ we have the equality $\iff \ a=mb$ si $\cos (C-\phi )=1\iff$

$m=\frac ab$ and $\phi =C$ and in this case we find the well-known identity $\sum a^2\left(b^2+c^2-a^2\right)=16S^2\ .$

Remark. $4r(5R-r)\le \sum bc\le 4(R+r)^2\ \Longleftrightarrow\ 16Rr-5r^2\le p^2\le 4R^2+4Rr+3r^2$ . Thus, exists the following inequalities: $36r^2\le 4S\sqrt 3\le\ 6r\sqrt [3]{4Rp^2}\le$

$4r(4R+r)\le 3\cdot\sqrt [3]{(abc)^2}\le$ $\sum a\sqrt {bc}\le \sum bc\le \frac 13\cdot \left(\sum a\right)^2\le \left\{\begin {array}{c} \frac 49\cdot (4R+r)^2\\\\ \sum a^2\end{array}\right\|$ and $\frac 13\cdot \left(a+b+c\right)^2\ \le$ $\left\{\begin {array}{c} \frac 49\cdot (4R+r)^2\\\\ a^2+b^2+c^2\end{array}\right\| \le$ $\ 4(2R^2+r^2)\ \le\ 9R^2\ .$

$\bigodot\ \boxed {\ a^2+b^2+c^2\ge 4S\sqrt 3+(a-b)^2+(b-c)^2+(c-a)^2\ }$ (Hadwiger-Finsler - see here).

Method I. $\sum a^2\ge \sum (b-c)^2+4S\sqrt 3\ \Longleftrightarrow\ \sum\left[a^2-(b-c)^2\right]\ \ge 4S\sqrt 3\ \Longleftrightarrow$ $\ \sum (p-b)(p-c)\ \ge\ S\sqrt 3\ .$ Denote $\left\|\ \begin{array}{c} (p-b)(p-c)=x\\\ (p-c)(p-a)=y\\\ (p-a)(p-b)=z\end{array}\ \right\|\ .$

But $S^2=yz+zx+xy\ .$ Hence $\sum x\ \ge \sqrt {3(xy+yz+zx)}\ \Longleftrightarrow\ \left(\sum x\right)^2\ \ge 3(xy+yz+zx)$ , what is true.

Method II. I"ll show that the Hadwiger-Finsler's inequality is equivalently with the well-known inequality $p\sqrt 3\le 4R+r\ .$ Indeed, $4S\sqrt 3+\sum (b-c)^2\ \le\ \sum a^2\ \Longleftrightarrow$

$4S\sqrt 3\ \le\ \sum \left[a^2-(b-c)^2\right]\ \Longleftrightarrow$ $\ 4S\sqrt 3\ \le\ 4\cdot\sum (p-b)(p-c)\ \Longleftrightarrow\ 4pr\sqrt 3\ \le\ 4r(4R+r)\ \Longleftrightarrow\ p\sqrt 3\ \le\ 4R+r\ .$

Method III. Prove easily that $2\cdot \sum (p-b)(p-c)=\sum a(p-a) .$ Thus, $\sum a^2-\sum (b-c)^2=$ $\ 4\cdot \sum (p-b)(p-c)=2\cdot\sum a(p-a)$ and the inequality

becomes $\sum a(p-a)\ \ge\ 2S\sqrt 3 .$ But $\sum a(p-a)\ge 3\cdot\sqrt [3]{abc(p-a)(p-b)(p-c)}=3\cdot \sqrt [3]{4Rp^2r^3}\ .$ I"ll show that $3\cdot\sqrt[3]{4Rp^2r^3}\ \ge\ 2S\sqrt 3\ .$

Indeed, this inequality is equivalently with $3R\sqrt 3\ \ge\ 2p .$ Therefore, we proved an stronger inequality, $\sum a^2\ \ge\ \sum (b-c)^2+6r\cdot\sqrt[3]{4Rp^2}\ .$

Method IV. $\sum a^2-\sum (b-c)^2=4\cdot\sum (p-b)(p-c)=4S\cdot \sum\tan\frac A2\ \ge 4S\sqrt 3$ because $\tan\frac x2\ : \ \left(0\ ,\ \pi\right)\rightarrow\mathbb R$ is convex and Jensen's $\implies \sum\tan\frac A2\ \ge\ \sqrt 3\ .$

$\bigodot$ A strengthening of the Hadwiger-Finsler's inequality: $\boxed {\ a^{2} + b^{2} + c^{2}\ge 4S \sqrt {3}+\frac {1}{2}\cdot \left(|a - b| + |b - c| + |c - a|\right)^{2}\ }$ .

Proof. Suppose w.l.o.g. $c\le b\le a$ . We need to prove that $a^2 + b^2 + c^2 - 2(a - c)^2\ge 4\sqrt {3}S$ . I"ll use $a = y+z$ , $b = z+x$ and $c = x+y$ ,

where $x\le y\le z\implies$ $y(x + y + z) + 3yz\ge 2\sqrt {3xyz(x + y + z)}$ which is true by AM-GM. Prove easily that

reverse of the Hadwiger-Finsley's inequality. i.e. $\boxed{4S\sqrt 3+\sum (b-c)^2\le a^2+b^2+c^2}\le 4S\sqrt 3+3\sum (b-c)^2$ .

Remark. Here Cezar Lupu & Cosmin Pohoata presented a "strengthening" of the Hadwiger-Finsler's inequality : $(G1)\ \sum a^2\ \ge\ \sum (b-c)^2+4S\cdot\sqrt {\frac {16R-5r}{4R+r}}\ \Longleftrightarrow$

$P\le \frac {4R+r}{p}$ , unde $P=\sqrt {\frac {16R-5r}{4R+r}}\ .$ Denote $K=\frac 1p\cdot \sqrt {4R^2+4Rr+3r^2}\ .$ Se observa ca $K\ge 1$ (Gerretsen) si $P\ge \sqrt 3\ .$ I"ll present a stronger inequality,

$(G2)\ \boxed {\ \sum a^2\ge \sum (b-c)^2+4S\cdot KP\ }\ \Longleftrightarrow\ \sqrt 3\le P\le \ KP\le \frac {4R+r}{p}\ .$ Indeed, $KP\le \frac {4R+r}{p}\ \Longleftrightarrow$ $4R+r\ge$ $\sqrt {4R^2+4Rr+3r^2}\cdot\sqrt {\frac {16R-5r}{4R+r}}$ $\Longleftrightarrow$

$\boxed {\ (4R^2+4Rr+3r^2)(16R-5r)\ \le\ (4R+r)^3\ }$ , i.e. $64R^3+44R^2r+28Rr^2-15r^3\le$ $64R^3+48R^2r+12Rr^2+r^3\ \Longleftrightarrow\ 4r(R-2r)^2\ge 0$ ,

what is true. An weak inequality: $9(R+r)\ \le\ 16R-5r\ \le\ \frac {(4R+r)^3}{4R^2+4Rr+3r^2}\ \le\ 9(2R-r)\ .$ Thus, exist the inequalities with the Hadwiger - Finsler's :

$\boxed {\ 4S\sqrt 3\ \le\ 4SP\ \le\ 4SKP\le\ 4r(4R+r)=\sum a^2-\sum (b-c)^2\ }\ \Longleftrightarrow$ $\sqrt 3\ \le\ P\ \le\ KP\ \le\ \frac {4R+r}{p}\ .$ Some applications of the inequality $(G2)$ which appear here.

$\boxed {A1}\ \ \sum\frac {a}{r_a}\ \ge\ 2KP\ \ge\ 2P\ \ge\ 2\sqrt 3\ .$

$\boxed {A2}\ \ \sum \tan\frac A2\ \ge\ KP\ \ge\ P\ \ge\ \sqrt 3\ .$

$\boxed {A3}\ \ \sum\frac {1}{h_ar_a}\ \ge\ \frac {KP}{S}\ \ge\ \frac PS\ \ge \frac {\sqrt 3}{pr}\ \ge\ \frac {9}{p^2}\ .$

$\boxed {A4}\ \ \sum\frac {1}{b+c-a}\ \ge \frac {KP}{2r}\ \ge\ \frac {P}{2r}\ \ge\ \frac {\sqrt 3}{2r}\ \ge\ \frac {9}{a+b+c}\ .$

$\boxed {A5}\ \ \sum\frac {a^2}{b+c-a}\ge 3R\cdot KP\ \ge$ $3R\cdot P\ge 3R\sqrt 3\ .$

$\boxed {A6}\ \ (a+b+c)^2\ \le\ (a+b+c)\cdot$ $\sum\frac {bc}{b+c-a}\ \le\ \frac {(5R-2r)(4R+r)^2}{2(2R-r)}\ .$

$\boxed {A7}\ \ \frac {a}{b+c-a}+\frac {b}{c+a-b}+\frac {c}{a+b-c}\ \ge\ \frac {b+c-a}{a}$ $+\frac {c+a-b}{b}+\frac {a+b-c}{c}+\frac {R-2r}{R}\cdot\frac {8R-r}{4(2R-r)}\ .$

$\boxed {A8}\ \ \sum\frac {1}{a(b+c-a)}\ \ge\ \frac {5R-2r}{8R^2r}$ $\Longleftrightarrow$ $\sum\frac {1}{1+\cos A}\ \ge\ 2+\frac {R-2r}{2R}$ $\Longleftrightarrow$ $\sum\frac {(a-b)(a-c)}{b+c-a}\ \ge\ p\cdot\frac {R-2r}{2R}\ge 0\ .$

I used the identities : $\sum\ \frac {1}{b+c-a}=\frac {4R+r}{2pr}\ ;\ \sum$ $\frac {b+c-a}{a}=\frac {p^2+r^2-8Rr}{2Rr}\ ;$ $\sum \frac {a}{b+c-a}=\frac {2R-r}{r}\ \ge\ 3\ ;\ \sum$ $a^2(p-b)(p-c)=4p^2r(R-r)\ ;$

$\sum\frac {a^2}{b+c-a}=\frac {2p(R-r)}{r}\ ;$ $\sum a(p-a)=2r(4R+r)\ ;\ \sum$ $\frac {(b+c)(2a-b-c)}{b+c-a}=\frac {2p(R-2r)}{r}\ ;\ \sum$ $\frac {(b+c)^2}{b+c-a}=8p+\sum\frac {(b+c)(2a-b-c)}{b+c-a}=$ $\frac {2p(R+2r)}{r}\ ;$

$\boxed {\sum a\cdot\sum\frac {bc}{b+c-a}=p^2+(4R+r)^2}\ .$

Remark. $4S\sqrt 3\ \le\ \left\{\begin{array}{c} 6r\sqrt [3]{4Rp^2}\\\\ 4SP\ \le\ 4SKP\end{array}\right\|\le\ 4r(4R+r)$ $\le 3\cdot\sqrt [3]{(abc)^2}\le$ $\sum a\sqrt {bc}\le$ $\sum bc\le\frac 13\cdot \left(\sum a\right)^2\le$ $\left\{ \begin {array}{c} \frac 49\cdot (4R+r)^2\\\\ a^2+b^2+c^2\end{array}\right\|\ \le\ \ 4(2R^2+r^2)\ \le\ 9R^2\ .$

$\bigodot$ Blundon's inequality. $\boxed {2R^{2} + 10Rr - r^{2} - 2(R - 2r)\sqrt {R(R - 2r)}\ \leq\ p^{2}\leq 2R^{2} + 10Rr - r^{2} + 2(R - 2r)\sqrt {R(R - 2r)}}\ .$

Proof. Denote $\left\|\begin{array}{c} \alpha\equiv \alpha (R,r,p)=2R^2+10Rr-r^2\\\\ \beta\equiv (R,r,p)=2(R-2r)\sqrt {R(R-2r)}\end{array}\right\|$ . Then the Blundon's inequality becomes $\boxed {\ \left|p^2-\alpha \right|\le\beta\ }\ .$ This remarkable inequality has proved by E. Rouche in 1851.

It is the fundamental inequality of $\triangle ABC$ . Let $\left\|\begin{array}{c} s_1=a+b+c=2p\\\\ s_2=ab+bc+ca=p^2+r^2+4Rr\\\\ s_3=abc=4Rpr\end{array}\right\|$ and symmetric homogenous polynomial $f(a,b,c)\equiv (a - b)^{2}(b - c)^{2}(c - a)^{2}=$

$s_{1}^{2}s_{2}^{2} - 4s_{2}^{3} - 4s_{1}^{3}s_{3} + 18s_{1}s_{2}s_{3} - 27s_{3}^{2} =$ $= - 4r^{2}[(p^{2} - 2R^{2} - 10Rr + r^{2})^{2} - 4r(R - 2r)^{3}]\ge 0$ $\implies$ $(p^{2} - 2R^{2} - 10Rp + r^{2})^{2}\leq 4R(R - 2r)^{3}$ .

Remark. Prove easily that $\left\|\begin{array}{c} ab+bc+ca=p^2+r(4R+r)\\\\ a^2+b^2+c^2=2\left[p^2-r(4R+r)\right]\\\\ a^3+b^3+c^3=2p\left[p^2-3r(2R+r)\right]\\\\ a^4+b^4+c^4=2\cdot\left(\sum bc\right)^2-16p^2r(2R+r)\end{array}\right\|$ si $\left\|\begin{array}{c} x=p-a>0\\\\ y=p-b>0\\\\\ z=p-c>0\end{array}\right\|$ $\Longleftrightarrow$ $\left\|\begin{array}{c} a=y+z\\\\ b=z+x\\\\ c=x+y\end{array}\right\|$ $\Longleftrightarrow$ $\left\|\begin{array}{c} x+y+z=p\\\\ xy+yz+zx=r(4R+r)\\\\ xyz=pr^2\end{array}\right\|\ ,$ i.e. $x$ , $y$ , $z$

verify the equation $t^3-pt^2+r(4R+r)t-pr^2=0\ .$ This chain of the equivalencies allows passing (directly and inverse) from one inequality $f(x,y,z)\ge 0$ with positive

$\{x,y,z\}$ to an inequality $h(a,b,c)\ge 0$ and finally to $g(R,r,p)\ge 0$ what can be proved with the fundamental Blundon's inequality. Thus can prove the following chain $:$

$r(16R-5r)\ \le\ \left|\begin{array}{c} \frac {2r(2R-r)(4R+r)}{R}\\\ r(16R - 5r) + \frac {r^2(R - 2r)}{R}\end{array}\right|\ \le$ $p^2\le \frac {R(4R+r)^2}{2(2R-r)}\le\ \left|\begin{array}{c} 4R(R+r)+3r^2\\\ \frac {R(4R+r)^2}{2(2R-r)}+\frac {3r^2(R-2r)}{2(2R-r)}\end{array}\right|\ .$ For $\left|\begin{array}{c} u\equiv u(R,r,p)\\\\ v\equiv v(R,r,p)\end{array}\right|$ say :

$\boxed {\ \begin{array}{ccccc} u\le p^2 & \Longleftrightarrow & u\le\alpha & \wedge & \alpha -u\ge \beta\\\\ p^2\le v & \Longleftrightarrow & v\ge \alpha & \wedge & v-\alpha \ge \beta \end{array}\ }$ , where $\left|\begin{array}{c} \alpha =2R^2+10Rr-r^2\\\\ \beta =2(R-2r)\sqrt {R(R-2r)}\\\\ \left|p^2-\alpha\right|\ \le\ \beta\end{array}\right|\ .$ Therefore, $\boxed {\ \begin{array}{c} \underline {\mathrm {Example\ I}}\blacktriangleright\ \ p^2\ \ge\ \frac {2r(2R-r)(4R+r)}{R}=u\\\\ \alpha -u=\left(2R^2+10Rr-r^2\right)-\frac {2r(2R-r)(4R+r)}{R}\\\\ \alpha -u=\frac 1R\cdot (R-2r)(2R^2-2Rr-r^2)\ge 0\\\\ \alpha -u\ge\beta\ \Longleftrightarrow\ 2R^2-2Rr-r^2\ge 2R\sqrt {R(R-2r)}\\\\ t=\frac Rr\ge 2\ \Longrightarrow\ \left(2t^2-2t-1\right)^2\ge 2t^3(t-2)\\\\ 4t+1\ge 0\ .\ \ \ \mathrm{O\ .\ K\ .}\end{array}\ }$ $\blacktriangleleft\blacktriangleright$ $\boxed {\ \begin{array}{c} \underline {\mathrm {Example\ II}}\blacktriangleright\ \ p^2\ \le\ \frac {R(4R+r)^2}{2(2R-r)}=v\\\\ v-\alpha=\frac {R(4R+r)^2}{2(2R-r)}-\left(2R^2+10Rr-r^2\right)\\\\ v-\alpha =\frac {(R-2r)\left(8R^2-12Rr+r^2\right)}{2(2R-r)}\ge 0\\\\ v-\alpha \ge \beta\ \Longleftrightarrow\ 8R^2-12Rr+r^2\ge 4(2R-r)\sqrt {R(R-2r)}\\\\ t=\frac Rr\ge 2\ \implies\ \left(8t^2-12t+1\right)^2\ge 16t(t-2)(2t-1)^2\\\\ (4t+1)^2\ge 0\ .\ \ \ \mathrm{O\ .\ K\ .}\end{array}\ }$ . In conclusion, in these examples proved : $\frac {2r(2R-r)(4R+r)}{R}\ \le\ p^2\ \le\ \frac {R(4R+r)^2}{2(2R-r)}\ .$

Example III. (Gazeta Matematica B, 12, 2003). For any $m\in (0,1]$ exists the bilateral inequality $u\ \le\ p^2\ \le\ v$ , where

$\left\|\begin{array}{c} u=\frac 1m\cdot\left[- (1 - m)^{2}R^{2} + 2(m^{2} + 5m + 2)Rr - (4 + m)r^{2}\right]=-mR(R-2r)+\alpha-\frac 1m\cdot (R-2r)^2\\\\ v=\frac 1m\cdot\left[(m + 1)^{2}R^{2} - 2(m^{2} - 5m + 2)Rr + (4 - m)r^{2}\right]=mR(R-2r)+\alpha +\frac 1m\cdot (R-2r)^2\end{array}\right\|\ .$

Proof. Observe that $u+v=2\alpha$ . Denote $\delta =\frac {v-u}{2}\ .$ Remain to show only $\delta\ \ge\ \beta\ .$ But $\delta=mR(R-2r)+\frac 1m\cdot (R-2r)^2=\frac 1m\cdot (R-2r)\left[\left(1+m^2\right)R-2r\right]\ge 0\ .$

In conclusion, $\delta\ge\beta\ \Longleftrightarrow\ \left(1+m^2\right)R-2r\ \ge\ 2m\sqrt {R(R-2r)}$ $\Longleftrightarrow$ $\left[\left(1+m^2\right)R-2r\right]^2-\left[2m\sqrt {R(R-2r)}\right]^2\ge 0$

$\Longleftrightarrow$ $\left(1-m^2\right)^2R^2-4\left(1-m^2\right)Rr+4r^2=\left[\left(1-m^2\right)R-2r\right]^2\ \ge\ 0$ , what is true.

Example IV. If numbers $x$ , $y$ , $z$ are pozitive, then exists the inequality $8(xyz)^2\ge \prod (y+z)\cdot\prod (y+z-x)\ \ (*)\ .$

Method 1. For positive $x$ , $y$ , $z$ exists $\triangle\ ABC$ so that $a+b+c=2p$ and $\boxed {\ \begin {array}{c} x=p-a\\\ y=p-b\\\ z=p-c\end{array}\ }\ .$ Observe that $y+z=a$ , $y+z-x=2a-p$ a.s.o. and the inequality $(*)$

becomes $8\left[(p-a)(p-b)(p-c)\right]^2+$ $abc\cdot (p-2a)(p-2b)(p-2c)\ge 0\ \ (1)\ .$ I"ll use the well-known inequalities $abc=4Rpr$ , $ab+bc+ca=p^2+r^2+4Rr$ and

$(p-a)(p-b)(p-c)=pr^2\ .$ Thus, the inequality $(1)$ is equivalently with $8\left(pr^2\right)^2+4Rpr\cdot\left[p^3-p^2\cdot 2(a+b+c)+p\cdot 4(ab+bc+ca)-8abc\right]\ge 0$

$\Longleftrightarrow$ $8p^2r^4+4Rpr\left[p^3-4p^3+4p(p^2+r^2+4Rr)-8\cdot 4Rpr\right]\ge 0$ $\Longleftrightarrow$ $2r^3+R\left[p^2-4r(4R-r)\right]\ge 0$ $\Longleftrightarrow$ $\boxed A\equiv 4r(4R-r)-\frac {2r^3}{R}\le p^2\ \ (2)\ .$ Therefore,

$(*)\ \Longleftrightarrow\ (1)\ \Longleftrightarrow\ (2)\ .$ I"ll use the fundamental inequality Blundon (only left) : $\boxed L\equiv 2R^2+10Rr-r^2-2(R-2r)\sqrt {R(R-2r)}\ \le\ p^2\ \ (3)\ .$ I"ll show

$A\le L\ .$ Indeed, $A\le L\ \Longleftrightarrow\ 4r(4R-r)-\frac {2r^3}{R}\le 2R^2+10Rr-r^2-2(R-2r)\sqrt {R(R-2r)}$ $\Longleftrightarrow$ $2(R-2r)\sqrt {R(R-2r)}\le 2R^2-6Rr+3r^2+\frac {2r^3}{R}$ $\Longleftrightarrow$

$2R(R-2r)\sqrt {R(R-2r)}\le 2R^3-6R^2r+3Rr^2+2r^3$ $\Longleftrightarrow$ $2R(R-2r)\sqrt {R(R-2r)}\le (R-2r)(\left(2R^2-2Rr-r^2\right)$ $\Longleftrightarrow$ $2R\sqrt {R(R-2r)}\le 2R^2-2Rr-r^2$ $\Longleftrightarrow$

$4R^3(R-2r)\le 4R^4-4R^2r(2R+r)+r^2\left(2R+r\right)^2$ , what is true. Thus, $A\le L\le p^2\ \implies\ A\le p^2$ , i.e. $(2)$ is true. Since $(*)\Longleftrightarrow (1)\Longleftrightarrow (2)$ , i.e. $(*)$ is true.

Method 2. For the positive $x$ , $y$ , $z$ exists the real $u$ , $v$ , $w$ so that $\boxed {\ \begin {array}{c} x=v+w\\\ y=w+u\\\ z=u+v\end{array}\ }\ .$ Observe that $y+z=2u+v+w$ , $y+z-x=2u$ a.s.o. and the inequality $(*)$ becomes

$uvw\prod (2u+v+w)\ \le\ \prod (v+w)^2\ .$ Thus, $\left|\begin{array}{c} 3uvw(u+v+w)\ \le\ (uv+vw+wu)^2\\\\ 27\prod(2u+v+w)\ \le\ 64(u+v+w)^3\\\\ 8(uv+vw+wu)(u+v+w)\ \le\ 9(u+v)(v+w)(w+u)\\\\ 8(uv+vw+wu)(u+v+w)\ \le\ 9(u+v)(v+w)(w+u)\end{array}\right|\ \bigodot\ \Longrightarrow$ $uvw\prod (2u+v+w)\ \le\ \prod (v+w)^2\ .$

Remark (Mateescu Constantin). Blundon's inequality is, in fact, a simple consequence of the triangle inequality !

$\boxed{\ \triangle\ ABC\ \implies\ 2R^2+10Rr-r^2-2(R-2r)\sqrt{R(R-2r)}\ \le\ s^2\ \le\ 2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)}\ }$

Short proof. I will use the well-known distances in a triangle $\boxed{\begin{array}{cccc} IN^2=s^2+5r^2-16Rr \\ \\ IO=\sqrt{R(R-2r)}\ \ ;\ \ ON=R-2r\end{array}}$ , where $I$ , $O$ and $N$ are the incenter , circumcenter and the Nagel's point

of the triangle $ABC$ . Now, one can easily show that Blundon's inequality reduces to $\boxed{\boxed{\ OI-ON\ \le\ IN\ \le\ OI+ON\ }}$ , which is just the triangle inequality applied in $\triangle\ NIO$ .

$\bigodot\ \boxed {\ x^2\left(b^2+c^2-a^2\right)+y^2\left(c^2+a^2-b^2\right)+z^2\left(a^2+b^2-c^2\right)\ge 16ST\ }$ (Neuberg-Pedoe's inequality - see here), where $(x,y,z)$ are the sides of $\triangle XYZ$ with the area $T\ .$

Proof. $x^2b^2+y^2a^2\ge 2xyab\ge 2xyab\cdot\cos (C-Z)\Longrightarrow$ $\ x^2b^2+y^2a^2-2xyab\cdot\cos C\cos Z\ge 2xyab\cdot\sin C\sin Z\Longrightarrow$ $2\left(x^2b^2+y^2a^2\right)-2xy\cos Z\cdot 2ab\cos C\ \ge$

$\ 4\cdot ab\sin C\cdot xy\sin Z$ $\Longrightarrow$ $2\left(x^2b^2+y^2a^2\right)-\left(x^2+y^2-z^2\right)\left(a^2+b^2-c^2\right)\ge 16ST$ $\Longrightarrow\sum x^2\left(b^2+c^2-a^2\right)\ \ge 16ST\ .$ We have the equality $\Longleftrightarrow$ $xb=ya$ and

$\cos (C-Z)=1$ $\Longleftrightarrow$ $\frac xa=\frac yb$ and $C=Z$ $\Longleftrightarrow$ $\triangle ABC\sim\triangle XYZ$ .

Remark. Let $(x,y,z)$ , $\left(\sqrt {\frac {a^2+x^2}{2}},\sqrt{\frac {b^2+y^2}{2}},\sqrt {\frac {c^2+z^2}{2}}\right)$ be sides for $\triangle XYZ$ , $\triangle MNP$ with the areas $T$ , $R$ respectively $\implies$ exists Oppenheim's inequality $\boxed {2R\ge S+T}$ .

Indeed, observe that $16S^2=\sum a^2\left(b^2+c^2-a^2\right)$ , $16T^2=\sum x^2\left(y^2+z^2-x^2\right)$ si $W\equiv\sum x^2\left(b^2+c^2-a^2\right)=\sum a^2\left(y^2+z^2-x^2\right)\ .$ Therefore,

$64R^2=\sum\left(a^2+x^2\right)\cdot\left[\left(b^2+c^2-a^2\right)+\left(y^2+z^2-x^2\right)\right]=$ $\sum a^2\left(b^2+c^2-a^2\right)+$ $\sum a^2\left(y^2+z^2-x^2\right)+$ $\sum x^2\left(b^2+c^2-a^2\right)+$ $\sum x^2\left(y^2+z^2-x^2\right)\ \implies$

$64R^2=16\left(S^2+T^2\right)+2W\ .$ Using the Neuberg-Pedoe's inequality $W\ge 16ST$ obtain that $4R^2\ge S^2+T^2+2ST\ \implies\ 2R\ \ge S+T\ .$ Prove similarly the extension of

the Oppenheim's inequality : if $m>0$ , $n>0$ , $m+n=1$ and $(x,y,z)$ , $\left(\ \sqrt {ma^2+nx^2}\ ,\ \sqrt {mb^2+ny^2}\ ,\ \sqrt {mc^2+nz^2}\right)$ are the lengths of the sides of $\triangle XYZ$ ,

$\triangle MNP$ with the areas $T$ , $R$ respectively, then there is the inequality $\boxed {\ R\ \ge\ mS+nT\ }\ .$

See here and here

This post has been edited 111 times. Last edited by Virgil Nicula, Feb 25, 2019, 7:27 PM

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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).

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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.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

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122. An identity for an equilateral triangle. Extension.

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

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119. Inspired by a Miculita's problem.

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107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

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104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

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94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

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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.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

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70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

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.

45. \cos B+\cos C=1 <==> nice property.

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.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

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).

Submit

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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

67. Remarkable algebraic/geometrical inequalities (1).

by Virgil Nicula, Jul 22, 2010, 11:12 PM

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