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2009 USAMO Problems/Problem 4

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Problem

For n \ge 2 let a_1, a_2, ..., a_n be positive real numbers such that

(a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2

Prove that max (a_1, a_2, ... ,a_n) \le  4 \text{min}\, (a_1, a_2, ... , a_n).

Solution

Assume without loss of generality that a_1 \geq a_2 \geq \cdots \geq a_n. Now we seek to prove that a_1 \le 4a_n.

By the Cauchy-Schwarz Inequality, \begin{align*}(a_n+a_2+ a_3 + ... +a_{n-1}+a_1)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) &\ge \left... Since a_1 \ge a_n, clearly (a_1 - {a_n \over 4}) > 0, dividing yields:

0 \ge (a_1 - 4a_n) \Longrightarrow 4a_n \ge a_1

as desired.

See Also

2009 USAMO (Problems • Resources: AoPS | ML)
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
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