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

(Created page with '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+..…')
(No difference)

Revision as of 23:40, 11 April 2011

Since each $a_{i}$ is positive, by Muirhead's inequality, $2(\sum a_{i}^2) \ge (\sum a)^2 \ge n$. Now we claim that $\frac{n}{2}> (\frac{1}{4})*(1+...\frac{1}{n})$

$n=1$, giving $1/2>1/4$ works, but we set the base case $n=2$, which gives $1>3/8$. Now assume that it works for $n$. By our assumption, now we must prove that, for $n+1$ case, which simplifies down to $1/2>\frac{1}{n+1}$, which is clearly true for $n>1$. So we are done.

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