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!

Another way.
Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(1+a^2)(1+b^2)(1+c^2)=((1-ab)^2+(a+b)^2)(1+c^2)=$$
$$=(1-ab-(a+b)c)^2+((1-ab)c+a+b)^2=$$
$$=(1-ab-ac-bc)^2+(a+b+c-abc)^2=4+(a+b+c-abc)^2.$$
Id est, it's enough to prove that
$$a+b+c-abc\geq2,$$ which is true by Maclaurin.
Indeed, let $a+b+c=3u$, $ab+ac+bc=v^2$, where $v>0$ and $abc=w^3$.
Thus, we need to prove that $$3u-w^3\geq2$$ or
$$3u^2v-2v^3\geq w^3,$$ which is true because $$u\geq v\geq w.$$
