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Let $a,b,c$ are positive real numbers such that, $a^2+b^2+c^2=3$. Prove that $$\frac{a^2}{a + b} + \frac{b^2}{b + c} + \frac{c^2}{c + a} \ge \frac{3}{2}$$

Drona
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    Hello,welcome to MSE.You will find that we will be more inclined to help you if you add your own thoughts and ideas to the body of your question.You will also find that people may consider it rude to have an imperative tone(prove that. . .).Also,adding the context and where you received the problem will probably prevent us from telling you something which you already know.Please edit the body of your question in order to add your own efforts. – rah4927 Mar 18 '14 at 17:04
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    Observe that $$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a},$$ so the inequality reduces to the symmetric, and rather easy, inequality $$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq 3.$$ – tipshoni Mar 18 '14 at 17:25
  • @tipshoni +1. Why not post it as an answer? The next step is proving $a^2+b^2\ge a+b$, right? How do we do that? – Zafer Cesur Mar 18 '14 at 19:14
  • How about Titu's lemma? – rah4927 Mar 19 '14 at 03:04
  • @ZaferCesur $a^2+b^2 \geq a+b$ is not necessarily true, take for example $a=b=\frac{1}{2}$ – Ewan Delanoy Mar 19 '14 at 06:26
  • Don't try to use the AM-GM inequality: $\sqrt[3]{\frac{a^2b^2c^2}{(a+b)(a+c)(b+c)}}$ can be lesser than $\frac 12$. – ajotatxe Mar 19 '14 at 08:55

1 Answers1

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As indicated in tipshoni’s comment, the inequality to be shown is equivalent to

$$ \sum_{cyc}\frac{a^2+b^2}{a+b} \geq 3 \tag{1} $$

This problem is question number 44 in chapter 1 of Vasile Cirtoaje’s book "Algebraic Inequalities" (available here).

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Ewan Delanoy
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