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let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$ show that $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$

This problem is from Secrets In Inequalities volume 1 page 30,example 1.24. the comment.the author this case is a bit more diffcult,But this author can't post solution.Thank you

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  • $\begingroup$ If $a=b=c=1$ then certainly $ab+bc+ac=3$ but it doesn't work for the inequality. Am I missing something here? $\endgroup$
    – imranfat
    May 17, 2014 at 17:00
  • $\begingroup$ I think the "second case is a bit more difficult" regarded the $\sqrt a + \sqrt b + \sqrt c $ inequality $\endgroup$
    – MT_
    May 17, 2014 at 22:40

1 Answer 1

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By AM-GM (with all sums being cyclic), $$\sum \frac{a^2}{a+2b^2} = \sum a - \sum \frac{2ab^2}{a+2b^2}\ge \sum a - \sum \frac{2ab^2}{3\sqrt[3]{ab^4}}= \sum a - \frac23\sum (ab)^{2/3}$$

By Power Mean Inequality, $$1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3} \implies \sum (ab)^{2/3} \le 3$$

It remains to show that $\sum a \ge 3$ which follows from the well known $$\left(\sum a\right)^2 \ge 3\sum ab = 9$$

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  • $\begingroup$ Shouldn't the $\frac{3}{2}$ in the root be a $\frac{2}{3}$? $\endgroup$
    – MT_
    May 17, 2014 at 22:37
  • $\begingroup$ i'm not very knowledgeable of this topic, so I just wanted to make sure before making any accusations :) $\endgroup$
    – MT_
    May 18, 2014 at 3:26
  • $\begingroup$ @MichaelT That's perfectly OK, and in fact thanks - my typo. $\endgroup$
    – Macavity
    May 18, 2014 at 3:29
  • $\begingroup$ how can you write $1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3}$??? $\endgroup$
    – Ishan
    Apr 25, 2020 at 4:57
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    $\begingroup$ @User88463 Read up on Power Means Inequality: en.wikipedia.org/wiki/…. The statement used above follows directly from $1> \frac23$. $\endgroup$
    – Macavity
    Apr 25, 2020 at 5:40

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