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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.

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  • $\begingroup$ Can down-voter explain us why did you make it? $\endgroup$ – Michael Rozenberg Jun 8 at 6:32
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    $\begingroup$ But you can up vote it Michael. @MichaelRozenberg $\endgroup$ – Aqua Jun 8 at 7:20
  • $\begingroup$ 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). $\endgroup$ – J.G. Jun 8 at 7:22
  • $\begingroup$ QM is quadratic mean? $\endgroup$ – Aqua Jun 8 at 13:35
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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?

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  • $\begingroup$ So you decided that this is unacceptable. $\endgroup$ – Aqua Jun 8 at 15:31

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