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

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
$$
Y \geq \sum_{cyc} \frac a {b + c} \geq \frac 3 2
$$
