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## IMC 2016 – Day 1 – Problem 3

Problem 3. Let ${n}$ be a positive integer. Also let ${a_1,a_2,...,a_n}$ and ${b_1,b_2,...,b_n}$ be reap numbers such that ${a_i+b_i >0}$ for ${i = 1,2,...,n}$. Prove that

$\displaystyle \sum_{i=1}^n \frac{a_ib_i -b_i^2}{a_i+b_i} \leq \frac{\sum_{i=1}^n a_i \cdot \sum_{i=1}^n b_i - \left(\sum_{i=1}^n b_i \right)^2 }{\sum_{i=1}^n (a_i+b_i)}.$

Sketch of proof: The strange condition that ${a_i+b_i>0}$ is too familiar to the hypothesis needed for the following variant of the Cauchy-Schwarz inequality to work: let ${x_1,...,x_n}$ be real numbers and ${y_1,...,y_n>0}$ be positive real numbers. Then

$\displaystyle \sum_{i=1}^n \frac{x_i^2}{y_i} \geq \frac{(x_1+...+x_n)^2}{y_1+...+y_n}$

Subtract ${\sum_{i=1}^n (a_i+b_i)/2}$ from each side of the members of the above inequalities. We obtain

$\displaystyle \sum_{i=1}^n \frac{-a_i^2/2 - 3b_i^2/2}{a_i+b_i} \leq \frac{-(\sum_{i=1}^n a_i)^2 /2 - 3(\sum_{i=1}^n b_i)^2/2}{\sum_{i=1}^n (a_i+b_i)}.$

Now we are done, since the resulting inequality is just an application of the above variant of the Cauchy-Schwarz inequality for ${x_i = a_i, y_i = a_i+b_i}$ and ${x_i = b_i, y_i =a_i+b_i}$.