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

Could you help me please with the following inequality

$a,b,c$ non-negative numbers such that: $a+b+c=3.$ Prove that:

$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}.$$

share|cite|improve this question
Have you tried something? Say, Lagrange multipliers, maybe? If you try to find the maximum of the function $f(a,b,c) = a \sqrt{2b+c^2} + \dots$, you might find as a result that $a=b=c=1$ is a maximum, thus giving you the answer. I haven't tried, but it's a suggestion. I am saying this because it is often nice to see that we don't get asked the question because the OP didn't want to try but rather because he is stuck and we are helping. I don't know what tools you have though, so I'm asking if you tried something beforehand. – Patrick Da Silva Sep 14 '12 at 19:10
You'd think that after asking ~40 questions about similar-looking inequalities, a user might start to see some commonality... – The Chaz 2.0 Sep 14 '12 at 19:37
up vote 0 down vote accepted


Using Cauchy-Schwarz we obtain :

$$\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c} \leq \sqrt{(a+b+c)\left(a^2b+b^2c+c^2a+2ab+2bc+2ca\right)}=\sqrt{3\left(\sum{2ab}+\sum{c^2a}\right)}.$$

we have to show that:

$$\sum{2ab}+\sum{c^2a} \leq 9 \Leftrightarrow$$

$$2\left(\sum a\right)\left(\sum ab\right)+3\sum c^{2}a\le\left(\sum a\right)^{3}\Leftrightarrow$$ $$ 2\left(\sum a\right)\left(\sum ab\right)+3\sum c^{2}a\le 2\left(\sum a\right)\left(\sum ab\right)+\left(\sum a\right)\left(\sum a^{2}\right)\Leftrightarrow$$ $$3\sum c^{2}a\le\left(\sum a\right)\left(\sum a^{2}\right)\Leftrightarrow$$ $$2\sum c^{2}a\le\sum c^{3}+\sum ca^{2}$$

and using $AM-GM$ inequality we proved the desired inequality.

Original source can be check here.

share|cite|improve this answer

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.