International Mathematics Competition
for University Students
2016

IMC2016: Day 1, Problem 3

3. Let $n$ be a positive integer. Also let $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ be real numbers such that $a_i+b_i>0$ for $i=1,2,\ldots,n$. Prove that $$\sum_{i=1}^n \frac{a_ib_i-b_i^2}{a_i+b_i}\leq \frac{\sum\limits_{i=1}^n a_i \cdot \sum\limits_{i=1}^n b_i -\left(\sum\limits_{i=1}^n b_i\right)^2}{\sum\limits_{i=1}^n (a_i+b_i)}.$$

Proposed by Daniel Strzelecki, Nicolaus Copernicus University in Torún, Poland

Hint: Use the following variant of the Cauchy-Schwarz inequality: $$\sum_{i=1}^n \frac{X_i^2}{Y_i} \ge \frac{(X_1+\ldots+X_n)^2}{Y_1+\ldots+Y_n} \quad (Y_1,\ldots,Y_n>0)$$

  Solution