Saudi Arabian Mathematical Competitions, 2012

From the AM-GM inequality we have
$$\frac{9}{x+y+z} - \frac{1}{xyz} \le \frac{3}{\sqrt[3]{xyz}} - \frac{1}{xyz}. \quad (1)$$
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 $x = y = z = 1$.