<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] By Holder $$3=x^n+y^n+z^n=\frac{1}{3^{n-1}}(1+1+1)^{n-1}(x^n+y^n+z^n)\geq\frac{1}{3^{n-1}}(x+y+z)^n,$$ which gives $$x+y+z\leq3.$$
Now, let $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=3$.
Thus, $k(a+b+c)\leq3$ gives $$0<k\leq1$$ and
$$\sum_{cyc}\frac{x}{x+yz}=\sum_{cyc}\frac{ka}{ka+k^2bc}=\sum_{cyc}\frac{a}{a+kbc}\geq\sum_{cyc}\frac{a}{a+bc}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{a+bc}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{a}{a+bc}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a-bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{a(a+b+c)-3bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a+3c)-(c-a)(a+3b)}{a+bc}\geq0$$ or 
$$\sum_{cyc}(a-b)\left(\frac{a+3c}{a+bc}-\frac{b+3c}{b+ac}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c(a+b+3c-3)}{(a+bc)(b+ac)}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c^2}{(a+bc)(b+ac)}\geq0$$ and we are done! 
</s>