<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] the pqr method:
Let $p = a+b+c, q = ab+bc+ca = 3, r = abc$.
Since $(a+b+c)^2 - 3(ab+bc+ca) = a^2+b^2+c^2 - ab-bc-ca \ge 0$, we have $p \ge 3$.
Since $ab+bc+ca \ge 3\sqrt[3]{ab \cdot bc \cdot ca} = 3\sqrt[3]{(abc)^2}$, we have $r\le 1$.
We have
\begin{align}
(1+a^2)(1+b^2)(1+c^2) - 8 &= a^2b^2c^2+a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2-7\\
&=  r^2 + q^2 - 2pr + p^2-2q - 7  \\
&= (p-r)^2 - 4 \\
&\ge (3-1)^2 - 4\\
&= 0.
\end{align}
We are done.
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