1. ## Inequality

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

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

2. Originally Posted by Sigyn3
Prove that for all positive real numbers $\displaystyle x,y,z$,

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

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

3. Another solution.

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

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

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

$\displaystyle 2(x+y+z)+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ \ge\ 9$
$\displaystyle \Rightarrow\ (x+y+z)\left(2+\frac{1}{xyz}\right)\ \ge\ 9$
suppose $\displaystyle \alpha, \beta, \gamma$ are positive real numbers with $\displaystyle \frac{\gamma^3}{\alpha^2 \beta} \leq \frac{729}{4}.$ prove that for all positive real numbers $\displaystyle x,y,z: \ \ \alpha + \frac{\beta}{xyz} \geq \frac{\gamma}{x+y+z}.$