Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers.

If you write LHS - RHS as a quadratic in $a$, then the discriminant is: $$D = 4(b+c)^2 - 4((b^2+1)(c^2+1)-2bc)(b^2+1)(c^2+1).$$ But $(b^2+1)(1+c^2)\geq (b+c)^2$ by C-S and $(b^2+1)(c^2+1)-2bc = b^2c^2+(b-c)^2+1\geq 1,$ so the discriminant is non-positive.

Proceeding along your approach:

It suffices to prove that $1 + p^2 - 2q + q^2 - 2pr + r^2 \ge 2q$.

Since $p^2 \ge 3q, q^2 \ge 3pr$, it suffices to prove that $1 + 3q - 2q + q^2 - \frac{2}{3}q^2 \ge 2q$
or $1 - q + \frac{q^2}{3} \ge 0$
or $\frac{1}{3}(q - \frac{3}{2})^2 + \frac{1}{4} \ge 0$ which is true. We are done.

Because $2(ab + bc + ca) \leqslant 2(|a||b| + |b||c| + |c||a|)$ and $$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) = (|a|^2 + 1)(|b| ^2 + 1)(|c| ^2 + 1),$$ so we need to prove the inequality when $ a,\,b,\,c \geqslant 0.$

Indeed, easy to check $3t^2 \geqslant 3t-1.$ Now, using the AM-GM we have $$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) \geqslant a^2 + b^2 + c^2 + 1 + a^2b^2 + b^2c^2 + c^2a^2$$ $$ \geqslant a^2+b^2+c^2+1+3\sqrt[3]{(abc)^4}$$ $$ \geqslant a^2+b^2+c^2+3\sqrt[3]{(abc)^2}.$$ Therefore we will show that $$a^2+b^2+c^2+3\sqrt[3]{(abc)^2} \geqslant 2(ab+bc+ca).$$ Which is very known (here, here).