<s>[INST] Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-

Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8  $$

Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
 [/INST] Here is another solution using AM-GM and Cauchy-Schwarz. Our goal is to prove that:
$$(1+a^2)(1+b^2)(1+c^2)\geq8$$
Rewriting the L.H.S. expression a little, we have:
$$(a^2+1)(1+b^2)(1+c^2) \ge (a+b)^2(1+c^2)=S,$$ where we have applied C-S to the first two terms in the product.
Hence it suffices for us to show that:
$$S=a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge (ab+bc+ca)^2-1$$
$$\iff a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge a^2b^2+2a^2bc+a^2c^2+2ab^2c+2abc^2+b^2c^2-1$$
$$\iff a^2+2ab+b^2 \ge a^2b^2+2a^2bc+2ab^2c-1$$
$$\iff (a+b)^2+1 \ge ab(ab+2ac+2bc)$$
$$\iff (a+b)^2+1 \ge ab(3+ac+bc) $$
$$\iff a^2+b^2+1 \ge ab(1+ac+bc)=ab(4-ab)$$
$$\iff a^2+b^2+a^2b^2+1 \ge 4ab$$
But the last inequality is trivially true by AM-GM; since:
$$a^2+b^2+a^2b^2+1 \ge 4 \sqrt[^4]{a^4b^4}=4ab,$$
and we are done.
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