Difference between revisions of "1998 USAMO Problems/Problem 3"
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By AM-GM, | By AM-GM, | ||
− | + | <cmath>\begin{align*} | |
− | + | \frac {1}{n}\sum_{j\neq i}{(1 - y_j)} &\geq \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}\\ | |
− | + | \frac {1 + y_i}{n} &\geq \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\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{(1 - y_i)}\\ | |
+ | \prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{n}\\ | ||
+ | \prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq n^{n + 1} | ||
+ | \end{align*}</cmath> | ||
− | Note that by the addition formula for tangents, < | + | Note that by the addition formula for tangents, <cmath>\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}</cmath>. |
− | So <math>\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}</math>, as desired. | + | So <math>\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}</math>, as desired. <math>\blacksquare</math> |
− | |||
− | <math>\ | ||
==See Also== | ==See Also== |
Revision as of 18:54, 10 May 2023
Problem
Let be real numbers in the interval
such that
Prove that
.
Solution
Let , where
. Then we have
By AM-GM,
Note that by the addition formula for tangents, .
So , as desired.
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.