Home > Inequalities, Olympiad, Problem Solving, Uncategorized > IMC 2016 – Day 1 – Problem 3

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}.

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