Find the minimum of $(1+a^2)(1+b^2)(1+c^2)$ where $a,b,c\geq 0$ 
Find the minimum of:
$$(1+a^2)(1+b^2)(1+c^2)  \ \ \  a,b,c\geq 0$$

Knowing that $$ab+bc+ac=27$$
I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$: 
$$\begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\\
&=26^2+(a+b+c-abc)^2\geq 26^2=676
\end{align}$$
where the condition $ab+bc+ac=27$ has been applied.
Hence the minimum is $676$ as soon as we show that there are $a,b,c\geq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?
