# 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?

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For $n\ge 2$, one can apply (n-1)-EV principle to reduce it to the case where two of the variables are equal, which reduces it to a single variable inequality. See Prop 1.2 of emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf for a statement of the principle. –  user27126 Nov 8 '13 at 6:03
For $n = 1$, homogenize the inequality, and expand. If you fix $x+y+z$ and $xy+yz+zx$, the expression is quadratic in $xyz$ with negative leading coefficient, which means its maximum attains when $xyz$ reaches its extrema. This implies that two of $x,y,z$ are equal, by uvw method: ohkawa.cc.it-hiroshima.ac.jp/AoPS.pdf/The%20uvw%20method.pdf In particular we can again assume two variables to be the same, which reduce this to a one-variable inequality. –  user27126 Nov 8 '13 at 6:18

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 $$\sum_{cyc}{x^{8/3}y^{8/3}z^{2/3}}\geq\sum_{cyc}{x^2y^2z^2}.$$
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Starting with x+y+z =3

From the relation between AM & GM:

(x+yz)/2 >= sqrt(xyz)

=> 2/(x+yz) <= 1/sqrt(xyz)

=> 2x/(x+yz) <= x/sqrt(xyz)

Similarly for others it can be shown that

2y/(y+xz) <= y/sqrt(xyz)

2z/(z+xy) <= z/sqrt(xyz)

Adding all the above inequations we get

2(x/(x+yz) + y/(y+zx) + z/(z+xy)) <= (x+y+z)/sqrt(xyz)

Also since x+y+z=3 i.e. (xyz)^1/3 <= (x+y+z)/3=1

=>(xyz)^1/3 <= 1

=>xyz <= 1

Also when a<=1, a^1/n <= a^1/n+1 which can be proved from the assumption that when a>1

a^n <= a^n+1

So (xyz)^1/2 <= (xyz)^1/3

(x+y+z)/(xyz)^1/2 >= (x+y+z)/(xyz)^1/3

I came upto this, someone try to reach the last step.

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No, you are lot of wrong. –  user94270 Nov 8 '13 at 8:46