Can we use QM-AM inequality to solve this?

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.

• Can down-voter explain us why did you make it? – Michael Rozenberg Jun 8 at 6:32
• But you can up vote it Michael. @MichaelRozenberg – Aqua Jun 8 at 7:20
• Shouldn't the right-hand side be $\frac12\sum_{i=1}^na_i$? This question isn't from IMO 1991 either (unless the I doesn't stand for international). – J.G. Jun 8 at 7:22
• QM is quadratic mean? – Aqua Jun 8 at 13:35

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 $$i$$, : $$\frac{a_i^2}{a_i+b_i}+ \frac{b_i^2}{a_i+b_i}= \frac{a_i^2+b_i^2}{a_i+b_i} \geq \frac{{1\over 2}(a_i+b_i)^2}{a_i+b_i} = {1\over 2}(a_i+b_i)$$