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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.

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  • $\begingroup$ since there is no equality condition for $a = b = c,$ using inequalities that requires this condition will not be useful for you. In other words, $p^2 = 3q$ if and only if $a = b= c.$ I would just write the whole thing as a quadratic in $a$ find it's minimum in terms of $b$ and $c$, then show that minimum is non-negative. $\endgroup$ – dezdichado Aug 25 at 4:30
  • $\begingroup$ @dezdichado Yes, I destroyed the equality case. Wouldn't you just write it in a quadratic of $a$ and show the discriminant is negative? Well, there is a sixth degree expression in the$ LHS$. $\endgroup$ – Book Of Flames Aug 25 at 4:37
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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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  • $\begingroup$ How did you get the second inequality? Please explain little more. $\endgroup$ – Book Of Flames Aug 25 at 12:17
  • $\begingroup$ @BookOfFlames Do you mean $p^2 - 3q = a^2 + b^2 + c^2 - ab - bc - ca \ge 0$? $\endgroup$ – River Li Aug 25 at 12:42
  • $\begingroup$ I meant how did you reach $1 + 3q - 2q + q^2 - \frac{2}{3}q^2 \ge 2q$ from $1 + p^2 - 2q + q^2 - 2pr + r^2 \ge 2q$ ? $\endgroup$ – Book Of Flames Aug 25 at 12:43
  • $\begingroup$ @BookOfFlames Since $p^2 \ge 3q$, $-2pr \ge -\frac{2}{3}q^2$ and $r^2 \ge 0$, we have $1 + p^2 - 2q + q^2 - 2pr + r^2 \ge 1 + 3q - 2q + q^2 - \frac{2}{3}q^2$. $\endgroup$ – River Li Aug 25 at 13:16
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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.

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  • $\begingroup$ Would it be useful to see the equality case? $\endgroup$ – Book Of Flames Aug 25 at 5:01
  • $\begingroup$ there is an equality case only when the quadratic has a real root. However, you can do more with the discriminant: the second inequality used above is equality only when $b = c = 0$, while the first C-S attains equality only when $b = c = 1.$ So the discriminant is actually always negative. $\endgroup$ – dezdichado Aug 25 at 5:05
  • $\begingroup$ I think the equality case actually exists. I checked my question twice, and it is correct. There you see $(1+a^2)(1+b^2)(1+c^2)\geq2(ab+bc+ca)$. The sign is not a strict inequality, but greater than or equal to. This says that equality condition holds at some $a,b,c$. So where does it hold? $\endgroup$ – Book Of Flames Aug 26 at 2:46
  • $\begingroup$ I literally proved to you in the previous comment that that was impossible. Choosing to ignore a completely valid proof is not gonna help you get better at mathematics. Besides, $\geq$ includes the possibility of $>$, so you cannot logically deduct the equality must be attained. I can write $6\geq 5$ and it will be completely fine. $\endgroup$ – dezdichado Aug 26 at 2:57
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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).

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  • $\begingroup$ How did $1+3\sqrt[3]{(abc)^4}\geq 3\sqrt[3]{(abc)^2}$ by AM-GM? $\endgroup$ – Book Of Flames Aug 25 at 4:44
  • $\begingroup$ It's easy $3t^2 \geqslant 3t - 1,$ with $t = \sqrt[3]{(abc)^2}$ $\endgroup$ – nguyenhuyen_ag Aug 25 at 4:49
  • $\begingroup$ @nguyenhuyen_ag For those of us not so experienced with inequalities, it would be instructive of you to supply a proof for the final inequality (the one you say is well-known). $\endgroup$ – ΑΘΩ Aug 25 at 5:03
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    $\begingroup$ @ΑΘΩ But it's indeed, well known. For the proof use Schur and Muirhead. $\endgroup$ – Michael Rozenberg Aug 25 at 5:07
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    $\begingroup$ @MichaelRozenberg Thank you as well for the reference. $\endgroup$ – ΑΘΩ Aug 25 at 5:11

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