<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] 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).
</s>