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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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    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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    @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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    @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