Problem 24: Another inequality

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Joined: Mar 10, 2015

by henderson, Jun 19, 2016, 8:51 AM

$\bf\color{red}Problem \ 24$ Let $a,b,c$ be non-negative real numbers, no two of which are zero. Prove that
$2\left(\frac{a^{3}}{b+c}+\frac{b^{3}}{c+a}+\frac{c^{3}}{a+b}\right)+(a+b+c)^{2}\geq 4(a^{2}+b^{2}+c^{2}).$ $\bf\color{red}My \ solution$ After some algebraic manipulations, we are able to write the given inequality as
$\color{red}S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2\geq 0,$ where
$\color{blue}S_a=\frac{b^2+c^2-a^2}{(c+a)(a+b)},$ $\color{blue}S_b=\frac{c^2+a^2-b^2}{(a+b)(b+c)}$ and $\color{blue} S_c=\frac{a^2+b^2-c^2}{(b+c)(c+a)}.$
Firstly, let's assume that $a\geq b\geq c.$ According to the $\bf\color{green}SOS \ theorem,$ it's enough to prove that $S_b,S_c$ and $b^2S_a+a^2S_b$ are all non-negative. It's obvious that $S_b$ and $S_c$ both are non-negative, since $c^2+a^2-b^2\geq c^2\geq 0$ and $a^2+b^2-c^2\geq a^2\geq 0.$
Now, we are left to prove that $b^2S_a+a^2S_b\geq 0$ $\bf\color{red}\iff$
$\frac{b^2(b^2+c^2-a^2)}{(c+a)(a+b)}+\frac{a^2(c^2+a^2-b^2)}{(a+b)(b+c)}\geq 0$ $\bf\color{red}\iff$
$\frac{b^5+b^3c^2-a^2b^3+b^4c+b^2c^3-b^2a^2c+a^2c^3+a^4c-a^2b^2c+a^3c^2+a^5-a^3b^2}{(a+b)(b+c)(c+a)}\geq 0$ $\bf\color{red}\iff$
$b^5+b^3c^2-a^2b^3+b^4c+b^2c^3-b^2a^2c+a^2c^3+a^4c-a^2b^2c+a^3c^2+a^5-a^3b^2\geq 0.$ Since $a^5+b^5\geq a^2b^3+a^3b^2,$ $a^4c+b^4c\geq 2a^2b^2c$ and the other terms are non-negative, we are done. $\square$

This post has been edited 5 times. Last edited by henderson, Sep 11, 2016, 7:37 PM

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Problem 24: Another inequality

by henderson, Jun 19, 2016, 8:51 AM

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