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1994 USAMO Problems/Problem 4: Difference between revisions

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== See Also ==
== See Also ==
{{USAMO newbox|year=1994|num-b=3|num-a=5}}
{{USAMO oldbox|year=1994|num-b=3|num-a=5}}

Revision as of 10:46, 12 April 2011

Problem 4

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

\[\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).\]

Solution

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, $1/2>\frac{1}{n+1}$, which is clearly true for $n>1$. So we are done.

See Also

Template:USAMO oldbox

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