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Prove the following inequality: for $a,b,c>0$ $$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$

What I tried is using substitution:

$p=a+b+c$

$q=ab+bc+ca$

$r=abc$

But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$

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You can. $a^2 (b+c)(c+a) + b^2 (a+b)(c+a) + c^2 (a+b)(b+c)$ is a symmetric polynomial. The reduction algorithm will give you an expression in terms of $p,q,r$. –  Cocopuffs Dec 25 '12 at 13:04

3 Answers 3

up vote 8 down vote accepted

By AM-GM inequality, $$\frac{a^2}{a+b} + \frac{a+b}{4} \ge a$$ Add up the similar inequalities obtained by cyclic substitution, you are done.

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1  
I think that its better to include the proof of this inequality. –  Amr Dec 25 '12 at 12:59
2  
There is nothing to include to prove. –  007resu Dec 25 '12 at 13:03
1  
+1. Very, very neat. –  user17762 Jan 20 '13 at 23:24

This is just Cauchy-Schwarz:

$$\left(\frac{a}{\sqrt{a+b}}\sqrt{a+b}+\frac{b}{\sqrt{b+c}}\sqrt{b+c}+\frac{c}{\sqrt{a+c}}\sqrt{a+c}\right)^2 \leq \left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} \right)\big( (a+b)+(a+c)+(b+c)\big)$$

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Cauchy-Swartz: $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+ \cdots +\frac{a_n^2}{b_n} \geq \frac{(a_1+a_2+ \cdots +a_n)^2}{b_1+b_2+ \cdot +b_n}$ So, $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

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