Team selection tests for JBMO 2018

First, by AM-GM we can show that $$x^2+y^2+1\ge xy+x+y\text{ for all }x,y,z\in \mathbb{R}.$$ Hence, $\sqrt{a^2+b^2+2}\ge \sqrt{|ab|+|a|+|b|+1}=\sqrt{(|a|+1)(|b|+1)}$. Construct similar inequalities and take the sum, we get $$\begin{array}{rl}& \sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2}\\ & \ge \sqrt{(|a|+1)(|b|+1)}+\sqrt{(|b|+1)(|c|+1)}+\sqrt{(|c|+1)(|a|+1)}.\end{array}$$ By AM-GM for three numbers, we get $$\begin{array}{rl}& \sqrt{(|a|+1)(|b|+1)}+\sqrt{(|b|+1)(|c|+1)}+\sqrt{(|c|+1)(|a|+1)}\\ & \ge 3\sqrt[3]{(|a|+1)(|b|+1)(|c|+1)}\\ & =3\sqrt[3]{|abc|+|ab|+|bc|+|ca|+|a|+|b|+|c|+1}\\ & \ge 3\sqrt[3]{abc+ab+bc+ca+a+b+c+1}\ge 3\sqrt[3]{7+1}=6\end{array}$$ By combining these two inequalities, we finish the proof. The equality occurs when $a=b=c=1$.