How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$

Question

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?

Answer 1

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

1. 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 $$\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2}\geq0.$$
2. 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 $$1\geq \frac{x^2+y^2+z^2}{3}.$$
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 $$\sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}\geq\sum_{cyc}{x^2y^2z^2}.$$

Answer 2

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.