# 1994 USAMO Problems/Problem 4

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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.