# Difference between revisions of "1994 USAMO Problems/Problem 4"

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+ | ==Problem 4== | ||

+ | Let <math>\, a_1, a_2, a_3, \ldots \,</math> be a sequence of positive real numbers satisfying <math>\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,</math> for all <math>\, n \geq 1</math>. Prove that, for all <math>\, n \geq 1, \,</math> | ||

+ | |||

+ | <cmath>\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).</cmath> | ||

+ | |||

+ | == Solution == | ||

Since each <math>a_{i}</math> is positive, by Muirhead's inequality, | Since each <math>a_{i}</math> is positive, by Muirhead's inequality, | ||

<math>2(\sum a_{i}^2) \ge (\sum a)^2 \ge n</math>. Now we claim that <math> \frac{n}{2}> frac{1}{4}(1+...\frac{1}{n)}</math> | <math>2(\sum a_{i}^2) \ge (\sum a)^2 \ge n</math>. Now we claim that <math> \frac{n}{2}> frac{1}{4}(1+...\frac{1}{n)}</math> | ||

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<math>n=1</math>, giving <math>1/2>1/4</math> works, but we set the base case <math>n=2</math>, which gives <math>1>3/8</math>. Now assume that it works for <math>n</math>. | <math>n=1</math>, giving <math>1/2>1/4</math> works, but we set the base case <math>n=2</math>, which gives <math>1>3/8</math>. Now assume that it works for <math>n</math>. | ||

By our assumption, now we must prove that, for <math>n+1</math> case, <math>1/2>\frac{1}{n+1}</math>, which is clearly true for <math>n>1</math>. So we are done. | By our assumption, now we must prove that, for <math>n+1</math> case, <math>1/2>\frac{1}{n+1}</math>, which is clearly true for <math>n>1</math>. So we are done. | ||

+ | |||

+ | == See Also == |

## Revision as of 11:45, 12 April 2011

## Problem 4

Let be a sequence of positive real numbers satisfying for all . Prove that, for all

## Solution

Since each is positive, by Muirhead's inequality, . Now we claim that

, giving works, but we set the base case , which gives . Now assume that it works for . By our assumption, now we must prove that, for case, , which is clearly true for . So we are done.