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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

3 Answers 3

up vote 3 down vote accepted

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 $AM - GM$:

$\frac{x+yz}2 \ge \sqrt{xyz}$

$\Rightarrow \frac 2{x+yz} \le \frac 1{\sqrt{xyz}}$

$\Rightarrow \frac{2x}{x+yz} \le \frac x{\sqrt{xyz}}$

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.

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No, you are lot of wrong. –  user94270 Nov 8 '13 at 8:46
    
Please try to use Latex formatting for a better readability –  pushpen.paul Nov 5 '14 at 18:01
    
@bAlex, You are wrong in the last line, $(xyz)^{\frac 12} \le (xyz)^{\frac 13}$ $\Rightarrow \frac 1{(xyz)^{\frac12}} \ge \frac 1{(xyz)^{\frac 13}}$ –  Subhadeep Dey Aug 14 at 5:37

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 $n=1$

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