<s>[INST] How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where real $x,y,z > 0 $ and $xy + yz + zx = 1$
Thank you for your help .
 [/INST] This is a sketch with some details I didn't want to fill in but they'll check out. Let $n>m\geq 0$ be fixed and
$$f(x,y,z) = \frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$$
we'll try to find the minimum under $xz+zy+yx=1$ using Lagrange multipliers. Let
$$ g(x,y,z) = xz + zy+ yx$$
Following the idea of the multipliers, we try to solve
$$ \nabla f = \lambda \nabla g$$
We see the first relationship as
$$ \frac{(n-m) x^{n-1}x^m+ n y^m x^{n-1}}{(x^m + y^m)^2}-m\frac{ x^{m-1} z^n }{(z^m + x^m)^2} = \lambda(z+y) $$
It's not so easy to see, but since all 3 equations are the same having $x,y,z$ mixed up, it means we must have $x=y=z= \frac{1}{\sqrt{3}}$ for the three equations to hold with
$$ \lambda = \frac{ n-m}{4} \left ( \frac{1}{\sqrt{3}} \right) ^{n-m-2} $$
Also, you'll see that this value is a minimum by checking the Hessian(or by a clever intuition), use $g(x,y,z)=1$ ,$x,y,z,>0$ and $n >m$. Thus
$$f(x,y,z) \Big | _g \geq \min_{x,y,z} f(x,y,z) \Big | _g = f \left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)= \frac{3}{2} \left ( \frac{1}{\sqrt{3}} \right)^{n-m} $$
Edit: It occurred to me I should say something about the max of $f(x,y,z) \big| _g$, you'll see it occurs at infinity since we may take $ y = \epsilon \approx 0$ then 
$$xz \approx 1 \implies x \approx \frac{1}{z}$$
So as $x \to \infty$ we see that
$$f(x,y,z) \Big |_g = \mathcal{O} (x^{n-m}) \to \infty $$
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