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 3 of 3: Simplify the lower bound
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∑i=1nai2ai+bi≥(∑ai)22∑ai=12∑ai.\sum_{i=1}^n\frac{a_i^2}{a_i+b_i}\ge\frac{(\sum a_i)^2}{2\sum a_i}=\frac12\sum a_i.
Detailed analysis

Substitute the denominator sum into the Cauchy-Schwarz estimate and cancel the positive common sum. This is exactly the required inequality.