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?

Using power means we have 
$x+y+z\le 3\cdot  {(\dfrac {x^n+y^n+z^n} {3})}^{\frac 1 n}=3$ 
Let $x\le y\le z$ then we have $z\ge 1$ so 
$\dfrac x {x+yz}+\dfrac y {y+zx} \ge \dfrac 2 {z+1} \Leftrightarrow \\$
$\dfrac {(z-1)z(x-y)^2} {(x+yz)(y+zx)(z+1)} \ge 0$  is true
Hence it is sufficies to prove
$\dfrac 2 {z+1} +\dfrac z {z+xy} \ge \dfrac 3 2$ 
Since from AmGm inequality we have $xy\le (\dfrac {x+y} 2)^2 \le (\dfrac {3-z} 2)^2$  it is sufficies to prove
$\dfrac 2 {z+1} +\dfrac {4z} {4z+(3-z)^2} \ge \dfrac 3 2 \Leftrightarrow \\$
$\dfrac {(3-z)(z-1)^2} {(z+1)(z^2-2z+9)} \ge 0$
which is obvious since $x+y+z\le 3\Rightarrow z\le 3$
equality holds if and only if $x=y=z=1$  or $z\rightarrow 3,x=y\rightarrow 0$
second case of equality is  valid only if $n=1$
