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 .
 A: Despite the attempts in the other answers and comments, and the two ``proofs'' given there, the inequality does not hold in general. We give two proofs for its failure:
First Proof (involving huge numbers): Set $x=3/7$, $y=4/7$, $z=37/49$. Then $xy+yz+zx=1$. For $m=7$, $n=8$ set $A=\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$ and $B=\frac{3}{2}(\frac{1}{\sqrt{3}})^{n-m}$. Then
\begin{equation}
A^2-B^2=-\frac{5464419604082977128654242570410694589510448147711713}{929486260504473222256638487813651283882185173994428050}<0.
\end{equation}
Second proof (more conceptual): Suppose that $0<x<y<z$. Then, with $n=m+1$,
\begin{equation}
\lim_{m\to\infty}
(\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m})
=
\lim_{m\to\infty}
(\frac{x}{1+(y/x)^m}+\frac{y}{1+(z/y)^m}+\frac{z}{1+(x/z)^m})=z
\end{equation}
Now set $x=\frac{1}{3}$, $y=\frac{2}{3}$, $z=\frac{7}{9}$. Then $xy+yz+zx=1$, but $z<\frac{3}{2}\frac{1}{\sqrt{3}}$. So the stated inequality does not hold for all sufficiently large $m$.
Remark: The same question was asked here at MathOverflow, and I gave a similar counterexample there.
A: 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
$$f(x,y,z) \Big |_g = \mathcal{O} (x^{n-m}) \to \infty $$
A: By the conditions, we have:
$$y =\frac{1-xz}{x+z}\ge0\quad\longrightarrow \quad xz\le1, z \le 1/x$$
Now, without loss of generality, assume $z\ge y\ge x$. We therefore have that the expression $f(x,y,z)$ in question satisfies:
$$f(x,y,z) \ge \frac{3}{2} \frac{x^n}{z^m}\ge \frac{3}{2}x^{n-m}$$
Now it remains to be shown that $x\ge 1/\sqrt{3}$. This is true since the minimal distance from the origin to the hyperboloid is exactly $1$, so that:
$$x^2+y^2+z^2 \ge 1 \quad\longrightarrow \quad 3x^2 \ge 1$$
Thus completing the proof.
