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!
 A: Another way.
By C-S and AM-GM we obtain: $$\prod_{cyc}(1+a^2)\geq(a+b)^2(1+c^2)=(a+b)^2+(ac+bc)^2\geq$$
$$\geq4ab+(3-ab)^2=a^2b^2-2ab+9=(ab-1)^2+8\geq8.$$
A: Another way.
We need to prove that $$\prod_{cyc}(3+3a^2)\geq8\cdot27$$ or
$$\prod_{cyc}(ab+ac+bc+3a^2)\geq8(ab+ac+bc)^3$$ or
$$\sum_{sym}(3a^4b^2+a^3b^3+3a^4bc-6a^3b^2c-a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0\succ(3,2,1),$ $(4,1,1)\succ(3,2,1)$ and $(3,3,0)\succ(2,2,2).$
A: 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.
A: 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.
A: 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 $$(x-1)^2(x^2+2x+9)\geq0.$$
A: 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 $$u\geq v\geq w.$$
A: 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.
