Problem 38: Beautiful inequality

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Joined: Mar 10, 2015

by henderson, Jul 11, 2016, 8:24 PM

$\color{red}\bf{Problem\ 38}$ Let $a,b,c$ be positive real numbers. Prove that $\frac{(a+b+c)^3}{3(a^2+b^2+c^2)} \geq \frac{ab(a+b)}{a^2+b^2}+\frac{bc(b+c)}{b^2+c^2}+\frac{ca(c+a)}{c^2+a^2}.$ $\color{red}\bf{My\ solution}$ $\frac{(a+b+c)^3}{3(a^2+b^2+c^2)} \geq \frac{ab(a+b)}{a^2+b^2}+\frac{bc(b+c)}{b^2+c^2}+\frac{ca(c+a)}{c^2+a^2}$ $\iff$
$\frac{(a+b+c)^3}{3(a^2+b^2+c^2)}-(a+b+c) \geq \left(\frac{ab(a+b)}{a^2+b^2}-\frac{a+b}{2}\right)+\left(\frac{bc(b+c)}{b^2+c^2}-\frac{b+c}{2}\right)+\left(\frac{ca(c+a)}{c^2+a^2}-\frac{c+a}{2}\right)$ $\iff$
$\frac{-(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)}{3(a^2+b^2+c^2)} \geq \frac{-(a+b)(a-b)^2}{2(a^2+b^2)}+\frac{-(b+c)(b-c)^2}{2(b^2+c^2)}+\frac{-(c+a)(c-a)^2}{2(c^2+a^2)}$ $\iff$
$\frac{2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)}{3(a^2+b^2+c^2)} \leq \frac{(a+b)(a-b)^2}{a^2+b^2}+\frac{(b+c)(b-c)^2}{b^2+c^2}+\frac{(c+a)(c-a)^2}{c^2+a^2}$ $\iff$
$2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)\leq \sum_{cyc}3(a+b)(a-b)^2+\sum_{cyc}\frac{3c^2(a+b)(a-b)^2}{a^2+b^2}$ $\iff$
$\sum_{cyc}2c(a-b)^2 \leq \sum_{cyc}(a+b)(a-b)^2+\sum_{cyc}\frac{3c^2(a+b)(a-b)^2}{a^2+b^2}.$ Here $\color{green}\textbf{AM-GM}$ helps:
$\sum_{cyc}(a+b)(a-b)^2+\sum_{cyc}\frac{3c^2(a+b)(a-b)^2}{a^2+b^2}\geq \sum_{cyc}2\sqrt{\frac{3c^2(a+b)^2(a-b)^4}{a^2+b^2}}\geq \sum_{cyc}2\sqrt{3c^2(a-b)^4}\geq \sum_{cyc}2\sqrt{c^2(a-b)^4}=\sum_{cyc}2c(a-b)^2,$ as desired.

This post has been edited 4 times. Last edited by henderson, Sep 11, 2016, 10:45 AM

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Problem 38: Beautiful inequality

by henderson, Jul 11, 2016, 8:24 PM

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