If $x,y,z>0$, $n,m\in N$, $n\ge m$ prove:$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$ 
Given positive numbers $x,y,z$ and $n,m$ positive integers with $n\ge m$ prove that: $$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$$

I tried doing it as follows:
If $x\ge y \ge z$, then:
$y+z\le z+x\le x+y$
Hence:
$x^n\ge y^n\ge z^n$ and $\frac{1}{(y+z)^m}+\frac{1}{(z+x)^m}\ge \frac{1}{(x+y)^m}$
From Tchebychev:
$3[\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}]\ge (x^n+y^n+z^n)[\frac{1}{(y+z)^m}+\frac{1}{(x+z)^m}+\frac{1}{(x+y)^m}]$
which is where I got stuck.
Could you please help me finish this question off?

Your idea of chebyshev is right we just have to tailor it with repect to what we have to prove
WLOG $x\ge y\ge z$ then $x^{n-m}\ge y^{n-m}\ge z^{n-m}$  and ${(\frac{x}{y+z})}^m\ge{( \frac{y}{x+z})^m}\ge {(\frac{z}{x+y})}^m $
By chebyshev: $$\sum_{cyc} \frac{x^n}{{(y+z)}^m}=\sum_{cyc} x^{n-m} {(\frac{x}{y+z})}^m\ge \frac{1}{3} \sum x^{n-m}\cdot \sum {(\frac{x}{y+z})}^m \tag S$$ Now by power mean inequality and Nesbitts inequality $$\sum {(\frac{x}{y+z})}^m\ge \dfrac{{(\sum \frac{x}{y+z})}^m}{3^{m-1}}\ge \frac{3}{2^m}$$  Use this in inequality (S)  directly to prove result
