MathLabs

Problem 3

Let a_1,a_2,...,a_n and b_1,b_2,...,b_n be positive real numbers with a_1+...+a_n=b_1+...+b_n. Prove that ∑i=1nai2ai+bi≥a1+⋯+an2\sum_{i=1}^n \frac{a_i^2}{a_i+b_i} \ge \frac{a_1+\cdots+a_n}{2}.
Step 1 of 3: Apply Cauchy-Schwarz
Previous stepNext step
(∑i=1nai2ai+bi)(∑i=1n(ai+bi))≥(∑i=1nai)2.\left(\sum_{i=1}^n\frac{a_i^2}{a_i+b_i}\right)\left(\sum_{i=1}^n(a_i+b_i)\right)\ge\left(\sum_{i=1}^n a_i\right)^2.
Detailed analysis

Use the inequality in Engel form on the positive denominators a_i+b_i. Positivity makes every division legitimate.