# Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$

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 :)

-
@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
@Iuli That pdf link is no longer valid. – pushpen.paul Nov 9 '14 at 19:08

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 $a, b, c\leq 1$.

-
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

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