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In the pdf which you can download here I found the following inequality which I can't solve it.

Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$

Thanks :)

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    @Sasha Thanks for editing my exercises :)2012-08-31
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    maybe taking square on both side would help2012-08-31
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    below of this page is solutions and hints,please check it2012-08-31
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    @dato can you give the link. I can't find the page with the solutions and hints. Thanks2012-08-31
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    @Iuli That pdf link is no longer valid.2014-11-09
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    See also: [Proving an Olympiad type inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\le\sqrt{3(\frac{a}b+\frac{b}c+\frac{c}a)}$](https://math.stackexchange.com/q/2240607)2017-04-18
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    Possible duplicate of [Proving an Olympiad type inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\le\sqrt{3(\frac{a}b+\frac{b}c+\frac{c}a)}$](https://math.stackexchange.com/questions/2240607/proving-an-olympiad-type-inequality-sqrt-frac2abc-sqrt-frac2bac)2017-04-18
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    In case somebody wonders why I suggested older question as a duplicate of the recently posted one, I will add link to [related discussion in chat](http://chat.stackexchange.com/transcript/message/36791190#36791190). And I'll add link to the discussion on the meta explaining that age is not the only thing to keep in mind when choosing duplicates: [Original post marked as duplicate](https://math.meta.stackexchange.com/q/16417#16418).2017-04-19

2 Answers 2

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Using cauchy Schwarz or AM-QM we have that $$LHS \leq \sqrt{3\sum_{cyc}\frac{2a}{b+c}}$$ It suffices to prove $$\sum_{cyc}\frac{2a}{b+c}\leq \sum_{cyc}\frac ab$$ By homogeneity we may suppose $a+b+c=1$.

Clearing out denominators this reduces to show $$2\sum_{cyc}a(a+b)(a+c)abc\leq \sum_{cyc}a^2c(a+b)(b+c)(c+a)$$ which is equivalent to $$0\leq\sum_{cyc} a^2c(a+c)(a+b)(b+c-2bc)$$ which is true by AM-GM and the fact that $a, b, c\leq 1$.

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    I have a question or two. 1) Why $a+b+c=1$ and 2) if we don't know that $a+b+c=1$ can we solve this inequality using another tricks? thanks :)2012-09-08
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    @luli to prove the inequality for $a$,$b$, and $c$ it suffices to prove it for $a'=\frac{a}{a+b+c}$, $b'=\frac{b}{a+b+c}$, and $c'=\frac{x}{a+b+c}$. Then $0 and $a'+b'+c'=1$.2016-11-13
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    The inequality $\sum\limits_{cyc}\frac{2a}{b+c}\leq \sum\limits_{cyc}\frac ab$ is wrong. Try $a=10$, $b=2$ and $c=1$.2017-03-08
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    This solution in incorrect...2017-04-19
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    @Iuli: Have you checked whether the claims that this answer is mistaken are true? If they are, are you going to unaccept this answer or not?2017-04-26
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    @AlexM. I do not understand your first question. Thanks!2017-04-26
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    @Iuli: This answer is mistaken, as explained by **Michael Rozenberg** above. Why don't you unaccept it?2017-04-26
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By C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}.$$ Thus, it remains to prove that $$\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}$$ or $$\sum_{cyc}(3a^6c^3+3a^5b^4+6a^5c^4+a^6b^2c+2a^6c^2b+4a^5b^3c+4a^5c^3b+7a^4b^4c+$$ $$+a^5b^2c^2-11a^4b^3c^2-12a^4c^3b^2-8a^3b^3c^3)\geq0,$$ which is obviously true.

Done!

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    @Iuli Solution of uforoboa is total wrong. What do you think about my solution?2017-03-08
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    Your solution seems right, and the other (accepted solution as of this time) is incorrect.2017-04-19
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    @MichaelRozenberg: Why is the last claim "obviously" true? It might be true, but why obviously? I'm staring at it and I don't see how to quickly prove it.2017-04-26
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    @Alex M Because any term with"+" $\geq$ than any term with "-" For example $\sum\limits_{cyc}a^5b^2c^2\geq\sum\limits_{cyc}a^4b^3c^2$ it's $\sum\limits_{cyc}a^4\geq\sum\limits_{cyc}a^3b$, which is true by Rearrangement.2017-04-26