<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] Another way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $v^2=1$ and we need to prove that:
$$a^2b^2c^2+\sum_{cyc}(a^2b^2+a^2)\geq7$$ or $f(w^3)\geq0,$ where $$f(w^3)=w^6+(9v^4-6uw^3)v^2+(9u^2-6v^2)v^4-7v^6.$$
But, $$f'(w^3)=2w^3-6uv^2w^3<0,$$ which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which by $uvw$ happens for equality case of two variables.
Since our inequality is symmetric, we can assume that $b=a$, which gives $c=\frac{3-a^2}{2a}.$
Can you end it now?
For $a^2=x$ I got $$(x-1)^2(x^2+2x+9)\geq0.$$
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