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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. – Harald Hanche-Olsen Sep 14 '12 at 15:42
up vote 3 down vote accepted

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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There is a problem in the second usage of rearrangement inequality. I mean if I am right you would like to say that $a(a+b)\leq b(b+c)\leq c(c+a)$ right? But how can you assume the inequality between the last two terms? I mean $b\leq c$ but $b+c\geq c+a$ – uforoboa Sep 14 '12 at 16:09
@Iuli's_inequality – uforoboa Sep 14 '12 at 16:13
@uforoboa The Rearrangement Inequality is applied to $\displaystyle \frac a {(b + c)^2} \leq \frac b {(c + a)^2} \leq \frac c {(a + b)^2}$ and $a + b \leq a + c \leq b + c$. – AlbertH Sep 14 '12 at 16:21
@Iuli For Nesbitt's Inequality you can also look at example 2.1.2 of "your book":…. – AlbertH Sep 14 '12 at 16:30
I applied RE to $\displaystyle \frac 1 {b + c} \leq \frac 1 {a + c} \leq \frac 1 {a + b}$ and $\displaystyle \frac {a(a + b)(a + c)} {(b + c)^2} \leq \frac {b(b + c)(b + a)} {(c + a)^2} \leq \frac {c(c + a)(c + b)} {(a + b)^2}$ obtaining $\displaystyle \frac 1 {b + c} \cdot \frac {a(a + b)(a + c)} {(b + c)^2} + \frac 1 {a + c} \cdot \frac {b(b + c)(b + a)} {(c + a)^2} + \frac 1 {a + b} \cdot \frac {c(c + a)(c + b)} {(a + b)^2} \geq \frac 1 {a + c} \cdot \frac {a(a + b)(a + c)} {(b + c)^2} + \frac 1 {a + b} \cdot \frac {b(b+c)(b+a)}{(c+a)^2} + \frac 1 {b+c}\cdot\frac {c(c+a)(c+b)}{(a+b)^2}$ – AlbertH Sep 14 '12 at 16:47

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