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

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