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For $a,b,c >0$ prove that :

$\frac{a}{(b+c)^4}+\frac{b}{(c+a)^4}+\frac{c}{(a+b)^4} \geq \frac{3}{2(a+b)(b+c)(c+a)}.$

I don't know how should I start. It is difficult for me because the denominators has the power equals with 4, I think it must decrease the rank of the power.

danke :)

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    Before you even get started, you might notice that both sides are homogeneous of degree $-3$ in the tuple $(a,b,c)$, by which I mean that if you replace $(a,b,c)$ by $(ta,tb,tc)$ then both sides are multiplied by $t^{-3}$. This already improves the chance that the inequality may be true. Also, you get equality for $a=b=c$, so the inequality is sharp.2012-09-14

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The inequality can be written $ X := \sum_{cyc} \frac {a(a + b)(a + c)}{(b + c)^3} \geq \frac 3 2 $ Let's suppose $a \leq b \leq c$. We have $ \frac 1 {b + c} \leq \frac 1 {a + c} \leq \frac 1 {a + b} \\ \frac {a(a + b)(a + c)} {(b + c)^2} \leq \frac {b(b + c)(b + a)} {(c + a)^2} \leq \frac {c(c + a)(c + b)} {(a + b)^2} $ Applying Rearrangement Inequality we get $ X \geq \sum_{cyc} \frac {a(a + b)}{(b + c)^2}=:Y $ Reapplyng Rearrangement Inequality and Nesbitt's Inequality we arrive to $ Y \geq \sum_{cyc} \frac a {b + c} \geq \frac 3 2 $

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    @AlbertH Yes, yes. I catch the idea how to apply . Thanks :)2012-09-14