Saudi Arabia Mathematical Competitions 2012

From the AM-GM inequality we have $$\frac{9}{x+y+z}-\frac{1}{xyz}\le \frac{3}{\sqrt[3]{xyz}}-\frac{1}{xyz}.(1)$$ Let $t=\frac{1}{\sqrt[3]{xyz}}$. According to inequality (1) it is sufficient to prove that $3t-t^3\le 2$ for $t>0$. The last inequality is equivalent to $t^3-3t+2\ge 0$, which is the same as $(t-1{)}^2(t+2)\ge 0$.

We have equality if and only if $t=1$ and we have equality in the AM-GM inequality. That is, $xyz=1$ and $x=y=z$, so $x=y=z=1$.