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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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math110
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  • If $a=b=c=1$ then certainly $ab+bc+ac=3$ but it doesn't work for the inequality. Am I missing something here? – imranfat May 17 '14 at 17:00
  • I think the "second case is a bit more difficult" regarded the $\sqrt a + \sqrt b + \sqrt c $ inequality – MT_ May 17 '14 at 22:40

1 Answers1

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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 $$\left(\sum a\right)^2 \ge 3\sum ab = 9$$

Macavity
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  • Shouldn't the $\frac{3}{2}$ in the root be a $\frac{2}{3}$? – MT_ May 17 '14 at 22:37
  • i'm not very knowledgeable of this topic, so I just wanted to make sure before making any accusations :) – MT_ May 18 '14 at 03:26
  • @MichaelT That's perfectly OK, and in fact thanks - my typo. – Macavity May 18 '14 at 03:29
  • how can you write $1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3}$??? – Ishan Apr 25 '20 at 04:57
  • i get by am-gm $ab+bc+ca /3$ > abc^ (2/3)..! – Ishan Apr 25 '20 at 04:58
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    @User88463 Read up on Power Means Inequality: https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality. The statement used above follows directly from $1> \frac23$. – Macavity Apr 25 '20 at 05:40
  • got it, thanks.. – Ishan Apr 25 '20 at 06:59
  • @Macavity can you pls give some hint on case 2 i.e when $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ .... – Ishan Apr 25 '20 at 07:15
  • i till now proved that $a+b+c$ has to be greater then or equal 3...but i am not able to show that in this case also $(a b)^{2 / 3}+(b c)^{2 / 3}+(c a)^{2 / 3} \leq 3$...any hint on which inequality should we use here ? – Ishan Apr 25 '20 at 07:18
  • I suggest you post it as a separate problem, rather than adding this on to comments in a post from 5-6 years ago. – Macavity Apr 25 '20 at 07:40
  • ok,posted you can check.. – Ishan Apr 25 '20 at 08:12