# 1998 USAMO Problems/Problem 3

## Problem

Let $a_0,\cdots a_n$ be real numbers in the interval $\left(0,\frac {\pi}{2}\right)$ such that $$\tan{\left(a_0 - \frac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1$$ Prove that $\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}$.

## Solution

Let $y_i = \tan{(a_i - \frac {\pi}{4})}$, where $0\le i\le n$. Then we have

• $y_0 + y_1 + \cdots + y_n\ge n - 1$
• $1 + y_i\ge \sum_{j\neq i}{(1 - y_j)}$
• $\frac {1 + y_i}{n}\ge \frac {1}{n}\sum_{j\neq i}{(1 - y_j)}$

By AM-GM,

• $\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
• $\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
• $\prod_{i = 0}^{n} {\frac{1 + y_i}{n}}\geq \prod_{i = 0}^{n} {\prod_{j\neq i} {(1 - y_j)}^{\frac {1}{n}}}$
• $= \prod_{i = 0}^n{(1 - y_i)}$
• $\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}$

Note that by the addition formula for tangents, $\tan{(a_i)} = \tan{[(a_i - \frac {\pi}{4}) + \frac {\pi}{4}]} = \frac {1 + \tan{(a_i - \frac {\pi}{4})}}{1 - \tan{(a_i - \frac {\pi}{4})}} = \frac {1 + y_i}{1 - y_i}$.

So $\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}} = \tan{(a_0)}\tan{(a_1)}\cdots \tan{(a_n)}\ge n^{n + 1}$, as desired. $\text{QED}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 