<s>[INST] Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+ca$ and $r = abc$.
$$ LHS = 1 + p^2 - 2q + q^2 - 2pr + r^2 \geq 2q \Rightarrow 1 + p^2 + q^2 + r^2 \geq 4q + 2pr $$
It's quite well-known that $p^2\geq 3q$ and $q^2\geq 3pr$. So,
$$ 1 + 3q + 3pr + r^2 \geq 4q + 2pr \Rightarrow 1 + pr + r^2 \geq q $$
But I don't know how to prove it. It can also be seen that $a\ge b\ge c$, but I can't exploit symmetry.
Any help is thankfully welcome.
 [/INST] 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.
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