1998 USAMO Problems/Problem 3

1998 USAMO Problem #1

Problem:

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

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,

$\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
$\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
$\prod_{i = 0}^n{\frac {1 + y_i}{n}}\ge \prod_{i = 0}^n{\prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$ (Error compiling LaTeX. Unknown error_msg)
$= \prod_{i = 0}^n{(1 - y_i)}$
$\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}$

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.

$\text{QED}$

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1998 USAMO Problems/Problem 3