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.
 A: 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.
A: 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.
A: 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).
