Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :)

share|cite|improve this question
@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@Iuli That pdf link is no longer valid.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
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 $a, b, c\leq 1$.

share|cite|improve this answer
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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.