<s>[INST] 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?
 [/INST] 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
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