### 1998 USAMO #3

Problem: Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that$\tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.$Prove that$\tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.$

Solution: Denote $x_i$ as $\tan(a_i - \frac{\pi}{4})$. We now write $\tan a_i$ as $\tan(a_i - 45^{\circ} + 45^{\circ})$. By the $\tan(a+b)$ formula, we know that this is equivalent to $$\frac{x_i + 1}{1 - x_i}.$$ If we let $y_i = \frac{1 - x_i}{2} \in (0,1)$, then it suffices to prove that if $\sum_0^n y_i \le 1$ and $y_i \ge 0$, then $\prod_{i=1}^n \left( \frac{1}{y_i}-1 \right) \ge n^{n+1}.$

It is easy to show this. Homogenizing the inequality, we must show

$$\prod_{i=1}^n \left( \frac{y_0+y_1+y_2+\dots+y_n}{y_i}-1 \right) \ge n^{n+1},$$

and by AM-GM, $$\frac{y_1+y_2+y_3+\dots+y_n}{y_0}\ge n \sqrt[n]{\frac{y_1y_2y_3\dots y_n}{y_1}},$$ from which we can simply cyclically product. $\square$