219. A very nice geometrical inequality.

by Virgil Nicula, Jan 30, 2011, 8:29 AM

Proposed problem. Let $ABC$ be a triangle which has most one angle exceeding $60^{\circ}$ . Prove that $\boxed{\ (ab+bc+ca)^2\ \ge\ 4S\sqrt 3\cdot (a^2+b^2+c^2)\ }$ .

Proof. We can assume (by symmetry) that $A\ge 60^{\circ}\ge B \ge C$ . Therefore, $\frac{b^2+c^2-a^2}{2bc}=\cos A \le \cos 60^{\circ}=\frac{1}{2} \ge \frac{a^2+c^2-b^2}{2ac} \ge \frac{b^2+c^2-a^2}{2bc}$ .

There are positive real numbers $x,y,z$ satisfying $a=y+z$ , $b=z+x$ , $c=x+y$ . So get $\frac{3yz}{x} \geq x+y+z \geq \frac{3zx}{y} \geq \frac{3xy}{z}$ . Rewriting the inequality in terms of

$x,y,z$ get $(x^2+y^2+z^2+3(xy+yz+zx))^2 \ge 8\sqrt{3xyz(x+y+z)}(x^2+y^2+z^2+xy+yz+zx)$ . By homogeneity can assume that

$\boxed{\ x+y+z=3\ }$ . So, we have to prove that $(9+xy+yz+zx)^2 \ge 24\sqrt{xyz}(9-xy-yz-zx)$ , where $x,y,z$ are positive reals satisfying

$\frac{yz}{x} \ge 1 \geq \frac{zx}{y} \ge \frac{xy}{z}$ and $x+y+z=3$ . By the condition, it is true that $(x-yz)(y-zx)(z-xy) \leq 0 \ (*)$ or equivalently

$x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2 \geq xyz(x^2+y^2+z^2+1)$ . But $x^2y^2+y^2z^2+z^2x^2 = (xy+yz+zx)^2-6xyz$ $\iff$ $\sum x^2=9-2\sum yz$ .

So, $(*)$ is equivalent to $(xy+yz+zx+xyz)^2 \ge 16xyz$ $\iff$ $xy+yz+zx \geq 4\sqrt{xyz}-xyz \ (**)$ . Now, our aim was to prove that

$(9+xy+yz+zx)^2 \ge 24(9-xy-yz-zx)\sqrt {xyz}$ and it is equivalent to $(9+xy+yz+zx)^2+24\sqrt{xyz}(xy+yz+zx) \geq 216\sqrt{xyz}$ .

Using $(**)$ get $(9+xy+yz+zx)^2+24\sqrt{xyz}(xy+yz+zx) \geq (9+4u-u^2)^2+24u(4u-u^2)$ where $\boxed{\ u=\sqrt{xyz}\ }$ . If we prove that

$(9+4u-u^2)^2+24u(4u-u^2) \ge 216u$ , then we are done. Thus, $(9+4u-u^2)^2+24u(4u-u^2) \ge 216u\Longleftrightarrow$

$(u-1)(u^3-31u^2+63u-81) \ge 0$ which is true since $u=\sqrt [3]{xyz}\le\frac {x+y+z}{3}=1$ and $u^2(u-31)+9(7u-9)<0$ .

Remark. In this case $(ab+bc+ca)^2\ \ge\ 4S\sqrt 3\cdot (a^2+b^2+c^2$ $)\ \ge\ 4S\sqrt 3\cdot 4S\sqrt 3\ \implies\ a^2+b^2+$ $c^2\ge\ ab+bc+ca\ge 4S\sqrt 3$ .
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:40 PM

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