Saudi Arabian Mathematical Competitions

Note that $3 a^{2}+6 b^{2} \geq (a+2 b)^{2}$ is true for all real numbers $a, b$. Indeed, after expanding and grouping we get $2 a^{2}-4 a b+2 b^{2} \geq 0$ which is equivalent to $2(a-b)^{2} \geq 0$. So with non-negative numbers $a, b, c$ we have
$$\begin{aligned} a \sqrt{3 a^{2}+6 b^{2}}+b \sqrt{3 b^{2}+6 c^{2}} & +c \sqrt{3 c^{2}+6 a^{2}} \\ \geq a(a+2 b)+b(b+2 c)+c(c+2 a) & =(a+b+c)^{2} . \end{aligned}$$
$\square$