<s>[INST] How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy} \right)(x^2+y^2+z^2+3xyz)\ge (x+y+z)^2$$
then we only prove
$$\dfrac{9}{x^2+y^2+z^2+3xyz}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow x^2+y^2+z^2+3xyz\le 6$$
Then I can't,and for $n$ how prove it?
 [/INST] Starting with $x+y+z =3$ 
From the relation between $AM - GM$: 
$\frac{x+yz}2 \ge \sqrt{xyz}$
$\Rightarrow \frac 2{x+yz} \le \frac 1{\sqrt{xyz}}$
$\Rightarrow \frac{2x}{x+yz} \le \frac x{\sqrt{xyz}}$
Similarly for others it can be shown that
$ \frac {2y}{y+xz} \le \frac y{\sqrt{xyz}}$
$\frac {2z}{z+xy} \le \frac z{\sqrt{xyz}}$
Adding all the above inequations we get
$2[\frac x{x+yz} + \frac y{y+zx} + \frac z{z+xy}] \le \frac {x+y+z}{\sqrt{xyz}}$
Also since $x+y+z=3$ i.e. $(xyz)^{\frac 13} \le \frac {x+y+z}3 = 1$
$\Rightarrow (xyz)^{\frac 13} \le 1$
$\Rightarrow xyz \le 1$
Also when $a \le 1$, $a^{\frac 1n} \le a^{\frac 1{n+1}}$ which can be proved from the assumption that when $a\gt 1$ 
$a^n \le a^{n+1}$
So $(xyz)^{\frac 12} \le (xyz)^{\frac 13}$
$\frac {x+y+z}{(xyz)^{\frac 12}} \le \frac {x+y+z}{(xyz)^{\frac 13}}$
I came upto this, someone try to reach the last step.
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