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Could you help me please with the following inequality

Let $a$, $b$ and $c$ be 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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    You'd think that after asking ~40 questions about similar-looking inequalities, a user might start to see some commonality...2012-09-14

2 Answers 2

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

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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Since $f(x)=\sqrt{x}$ is a concave function, by Jensen we obtain: $\sum_{cyc}a\sqrt{2b+c^2}=3\sum_{cyc}\frac{a}{3}\sqrt{2b+c^2}\leq3\sqrt{\sum_{cyc}\frac{a}{3}(2b+c^2)}=\sqrt{3\sum_{cyc}(2ab+a^2b)}=$ $=\sqrt{3(9-\sum_{cyc}(a^2-a^2b)}=\sqrt{27-\left((a+b+c)(a^2+b^2+c^2)-3\sum_{cyc}a^2b\right)}=$ $=\sqrt{27-\sum_{cyc}(a^3-2a^2b+a^2c)}=\sqrt{27-\sum_{cyc}(a^3-2a^2b+b^2a)}=\sqrt{27-\sum_{cyc}a(a-b)^2}\leq3\sqrt3$