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.

$\text{QED}$

1998 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
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1998 USAMO Problems/Problem 3

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