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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 :) –  Iuli Aug 31 '12 at 5:21
    
maybe taking square on both side would help –  dato datuashvili Aug 31 '12 at 9:09
    
below of this page is solutions and hints,please check it –  dato datuashvili Aug 31 '12 at 9:26
    
@dato can you give the link. I can't find the page with the solutions and hints. Thanks –  Iuli Aug 31 '12 at 9:29

2 Answers 2

up vote 5 down vote accepted

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 :) –  Iuli Sep 8 '12 at 14:31

$$\sqrt{3\Big(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\Big)} \geq 3$$ by AM-GM

and $$\sqrt{\frac{2a}{b+c}} +\sqrt{\frac{2b}{a+c}} + \sqrt{\frac{2c}{a+b}} \leq 3$$ because $$\sqrt{\frac{2a}{b+c}}\leq \sqrt{\frac{a}{\sqrt{bc}}}$$ and AM-GM

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The second line is obviously false: try $a=b=c$. –  Brian M. Scott Aug 31 '12 at 22:44
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Why is the second line false if a = b = c? –  frogeyedpeas Jul 20 '13 at 2:11

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