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&gt;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 &gt; 0 $ and $xy + yz + zx = 1$

Thank you for your help .
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}&lt;0.
\end{equation}

Second proof (more conceptual): Suppose that $0&lt;x&lt;y&lt;z$. 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 $z&lt;\frac{3}{2}\frac{1}{\sqrt{3}}$. 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.
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.