By AM-QM $$LHS=\sum\limits_{cyc}\sqrt{a^2+b^2+1^2+1^2}\ge \sum\limits_{cyc}\frac{|a|+|b|+1+1}{2}=|a|+|b|+|c|+3.$$Thus it suffices to show that $|a|+|b|+|c|\ge 3$ follows from the given condition $(1+a)(1+b)(1+c)\ge 8$. Note that $$8\le (1+a)(1+b)(1+c)\le |1+a|\cdot|1+b|\cdot|1+c|\le \left[\frac{|1+a|+|1+b|+|1+c|}{3}\right]^3\le \left[\frac{|a|+|b|+|c|+3}{3}\right]^3$$whence immediately $|a|+|b|+|c|\ge 3$.

JBMO2020 wrote: Let $a, b, c$ be reals which satisfy $a+b+c+ab+bc+ac+abc=>7$, prove that $$\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2}=>6$$ https://artofproblemsolving.com/community/c6h1679728p10706236

Note that $$(a-b)^2 + (a-1)^2 + (b-1)^2 \ge 0 \Longleftrightarrow 2a^2 + 2b^2 -2ab-2a-2b+2 \ge 0 \Longleftrightarrow a^2 + b^2 -ab - a -b + 1 \ge 0 \Longleftrightarrow a^2 + b^2 + 2 \ge ab + a + b + 1 \Longleftrightarrow \sqrt{a^2 + b^2 + 2} \ge \sqrt{(a+1)(b+1)}$$Therefore, $$\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge \sqrt{(a+1)(b+1)} + \sqrt{(a+1)(c+1)} + \sqrt{(b+1)(c+1)} \ge 3 \sqrt[3]{(a+1)(b+1)(c+1)} \ge 3 \sqrt[3]{8} \ge 6$$as desired.