Difference between revisions of "2009 USAMO Problems/Problem 4"
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=&\left(n+\frac{1}{2}\right)^2 \\ | =&\left(n+\frac{1}{2}\right)^2 \\ | ||
\ge& (a_1+a_2+ a_3 + ... +a_{n})\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right).\end{align*}</cmath> | \ge& (a_1+a_2+ a_3 + ... +a_{n})\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right).\end{align*}</cmath> | ||
+ | (Note that <math>n-2 \ge 0</math> since <math>n \ge 2</math> as given!) | ||
This implies that <math>3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1</math> as desired. | This implies that <math>3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1</math> as desired. | ||
Revision as of 08:48, 29 December 2020
Problem
For let , , ..., be positive real numbers such that
Prove that .
Solution
Assume without loss of generality that . Now we seek to prove that .
By the Cauchy-Schwarz Inequality, Since , clearly , dividing yields:
as desired.
Alternative Solution (by Deng Tianle, username: Leole) Assume without loss of generality that . Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, (Note that since as given!) This implies that as desired.
See Also
2009 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.