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 .

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.
