SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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