There are two sequences (${a_1,a_2,a_3,....,a_n })$ and $( {b_1,b_2,b_3,....,b_n})$ such that $$\sum_{i=1}^n a_i = \sum_{i=1}^n b_i$$
Prove that: $$\sum_{i=1}^n \frac{a_i^2}{a_i+b_i} \ge \frac{1}{2} \sum_{i=1}^n a_i$$
P.S I can do it with the Cauchy-Schwarz inequality in the Engel form. But can you do it with QM-AM Inequality? I saw somebody do it here. I cannot understand it.
Note that for each $i$ we have $$\frac{a_i^2}{a_i+b_i}- \frac{b_i^2}{a_i+b_i} = a_i-b_i$$
so $$\sum_{i=1}^n \frac{a_i^2}{a_i+b_i} = \sum_{i=1}^n \frac{b_i^2}{a_i+b_i}$$
By Qm-Am inequality we have for each $i$, : $$\frac{a_i^2}{a_i+b_i}+ \frac{b_i^2}{a_i+b_i}= \frac{a_i^2+b_i^2}{a_i+b_i} \geq \frac{{1\over 2}(a_i+b_i)^2}{a_i+b_i} = {1\over 2}(a_i+b_i)$$
Does this help?
