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?
 A: For the proof in case $n\geq 2$, I found Muirhead's inequality very useful. 


*

*Expand the inequality. You have $$3xyz+\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^3yz}-3x^2y^2z^2 \geq 0$$, or stated another way, $$xyz(3-x^2+y^2+z^2)+(\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2})\geq0.$$
Therefore it is enough to prove that $$x^2+y^2+z^2\leq3$$ and $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq0.$$

*Showing $x^2+y^2+z^2\leq3$ is easy. We have $x^n+y^n+z^n=3$ for some $n\geq2$. Think about a function $f(x)=x^{n/2}$. It is convex because $n/2\geq1$, so we can use Jensen's inequality. Therefore, $$1=\frac{x^n+y^n+z^n}{3}=\frac{f(x^2)+f(y^2)+f(z^2)}{3}\geq f(\frac{x^2+y^2+z^2}{3})$$, and this implies $$1\geq \frac{x^2+y^2+z^2}{3}.$$

*Now we have to prove $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq0.$$ Notice that $1=\frac{x^n+y^n+z^n}{3}\geq x^{n/3}y^{n/3}z^{n/3}$, so $1 \geq xyz$. Because $xyz\leq1$, we have $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq \sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}-\sum_{cyc}{x^2y^2z^2}.$$ Here, we're done because Muirhead's inequality says $$\sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}\geq\sum_{cyc}{x^2y^2z^2}.$$

A: 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.
A: 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 $n=1$
A: 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! 
