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

Solution. By applying the identity $$ \frac{XY-Y^2}{X+Y} = Y - \frac{2Y^2}{X+Y} $$ with $X=a_i$ and $Y=b_i$ to the terms in the LHS and $X=\sum\limits_{i=1}^n a_i$ and $Y=\sum\limits_{i=1}^n b_i$ to the RHS, $$ LHS = \sum_{i=1}^n \frac{a_ib_i-b_i^2}{a_i+b_i} = \sum_{i=1}^n \left(b_i-\frac{2b_i^2}{a_i+b_i}\right) = \sum_{i=1}^n b_i -2 \sum_{i=1}^n \frac{b_i^2}{a_i+b_i}, $$ $$ RHS = \frac{\sum\limits_{i=1}^n a_i \cdot \sum\limits_{i=1}^n b_i - \left(\sum\limits_{i=1}^n b_i\right)^2}{ \sum\limits_{i=1}^n a_i + \sum\limits_{i=1}^n b_i} = \sum\limits_{i=1}^n b_i - 2\frac{ \left(\sum\limits_{i=1}^n b_i\right)^2}{\sum\limits_{i=1}^n (a_i+b_i)}. $$ Therefore, the statement is equivalent with $$ \sum\limits_{i=1}^n \frac{b_i^2}{a_i+b_i} \ge \frac{\Big(\sum\limits_{i=1}^n b_i\Big)^2}{\sum\limits_{i=1}^n (a_i+b_i)}, $$ which is the same as the well-known variant of the Cauchy-Schwarz inequality, $$ \sum_{i=1}^n \frac{X_i^2}{Y_i} \ge \frac{(X_1+\ldots+X_n)^2}{Y_1+\ldots+Y_n} \quad (Y_1,\ldots,Y_n>0) $$ with $X_i=b_i$ and $Y_i=a_i+b_i$.