Inequality

**Sigyn3** · August 26th 2008, 02:33 AM

Prove that for all positive real numbers $x,y,z$ ,

$2+\frac{1}{xyz}\ \ge\ \frac{9}{x+y+z}$ .

**NonCommAlg** · August 26th 2008, 04:22 AM

Originally Posted by Sigyn3

Prove that for all positive real numbers $x,y,z$ ,

$2+\frac{1}{xyz}\ \ge\ \frac{9}{x+y+z}$ .

by AM-GM we have $x+y+z \geq 3 \sqrt[3]{xyz}.$ thus: $\frac{9}{x+y+z} \leq \frac{3}{\sqrt[3]{xyz}}.$ so we only need to prove that $\frac{3}{\sqrt[3]{xyz}} \leq 2 + \frac{1}{xyz}. \ \ \ \ \ \ (1)$

let $xyz=\frac{1}{a^3}, \ a > 0.$ then (1) becomes $a^3 - 3a + 2 \geq 0,$ which is obviously true because $a^3 - 3a + 2 =(a-1)^2(a+2). \ \ \ \square$

**JaneBennet** · August 26th 2008, 03:33 PM

Another solution.

By AM–GM, $x+y+\frac{1}{xy}\ \ge3\ \sqrt[3]{(x)(y)\left(\frac{1}{xy}\right)}\ =\ 3$ .

Similarly $y+z+\frac{1}{yz}\ \ge\ 3$ .

And $z+x+\frac{1}{zx}\ \ge\ 3$ .

Adding up,

$2(x+y+z)+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ \ge\ 9$

$\Rightarrow\ (x+y+z)\left(2+\frac{1}{xyz}\right)\ \ge\ 9$

**NonCommAlg** · August 26th 2008, 08:23 PM

here's a general form of the inequality that i just made it up:

suppose $\alpha, \beta, \gamma$ are positive real numbers with $\frac{\gamma^3}{\alpha^2 \beta} \leq \frac{729}{4}.$ prove that for all positive real numbers $x,y,z: \ \ \alpha + \frac{\beta}{xyz} \geq \frac{\gamma}{x+y+z}.$

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