Team Selection Tests for JBMO, 2018

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