IMC 2016 · Day 1 · P3

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, $$\mathrm{LHS} = \sum_{i=1}^{n} \frac{a_i b_i - b_i^2}{a_i + b_i} = \sum_{i=1}^{n} \left( b_i - \frac{2 b_i^2}{a_i + b_i} \right) = \sum_{i=1}^{n} b_i - 2 \sum_{i=1}^{n} \frac{b_i^2}{a_i + b_i},$$ $$\mathrm{RHS} = \frac{\sum\limits_{i=1}^{n} a_i \cdot \sum\limits_{i=1}^{n} b_i - \left( \sum\limits_{i=1}^{n} b_i \right)^2} {\sum\limits_{i=1}^{n} a_i + \sum\limits_{i=1}^{n} b_i} = \sum_{i=1}^{n} b_i - 2 \frac{\left( \sum\limits_{i=1}^{n} b_i \right)^2} {\sum\limits_{i=1}^{n} (a_i + b_i)}.$$ Therefore, the statement is equivalent with $$\sum_{i=1}^{n} \frac{b_i^2}{a_i + b_i} \ge \frac{\left( \sum\limits_{i=1}^{n} b_i \right)^2} {\sum\limits_{i=1}^{n} (a_i + b_i)},$$ which is the same as the well-known variant of the Cauchy-Schwarz inequality, $$\sum_{i=1}^{n} \frac{X_i^2}{Y_i} \ge \frac{(X_1 + \dots + X_n)^2}{Y_1 + \dots + Y_n} \qquad (Y_1, \dots, Y_n > 0)$$ with $X_i = b_i$ and $Y_i = a_i + b_i$.