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Could you help me please with the following inequality

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

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

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5  
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
7  
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

1 Answer 1

up vote 0 down vote accepted

$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2}=\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c}$$

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.

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