1991 APMO Problems/Problem 3

Problem

Let $a_1$ , $a_2$ , $\cdots$ , $a_n$ , $b_1$ , $b_2$ , $\cdots$ , $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$ . Show that $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}$

Solution

By Cauchy-Schwarz, $\left(\sum \dfrac{a_i^2}{a_i+b_i}\right)\left(\sum a_i+b_i\right)\geq \left(\sum a_i\right) ^2$ , so $\sum \dfrac{a_i^2}{a_i+b_i}\geq \dfrac{\left(\sum a_i\right) ^2}{\sum (a_i+b_i)}=\dfrac{\sum a_i}{2} since$ \sum a_i=\sum b_i$.

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1991 APMO Problems/Problem 3

Problem

Solution