<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] Let $a=\sqrt3\tan\alpha$, $b=\sqrt3\tan\beta$ and $c=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right)$.
Thus, $$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we need to prove that:
$$\ln\left((1+a^2)(1+b^2)(1+c^2)\right)\geq3\ln2$$ or
$$\ln(1+a^2)-\ln2+\ln(1+b^2)-\ln2+\ln(1+c^2)-\ln2\geq0$$ or
$$\sum_{cyc}(\ln(1+a^2)-\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+3\tan^2\alpha\right)-\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln\left(1+3\tan^2\alpha\right)-\ln2-\frac{4}{\sqrt3}\left(\alpha-\frac{\pi}{6}\right)\right)\geq0,$$ which is true because easy to show that for any $x\in\left(0,\frac{\pi}{2}\right)$ we have $f(x)\geq0,$ where
$$f(x)=\ln\left(1+3\tan^2x\right)-\ln2-\frac{4}{\sqrt3}\left(x-\frac{\pi}{6}\right).$$
Also, we see that $$f''(x)=\frac{6(4\cos^4x-6\cos^2x+3)}{\cos^2x(2-\cos2x)^2}>0$$ and our inequality follows from Jensen.
</s>