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 2 of 3: Use the equal-sum hypothesis
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∑i=1n(ai+bi)=∑ai+∑bi=2∑ai.\sum_{i=1}^n(a_i+b_i)=\sum a_i+\sum b_i=2\sum a_i.
Detailed analysis

The hypothesis says the two sums are equal, so the sum of all denominators is twice the common sum.