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

Question

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 :)

Answer 1

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?

Answer 2

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial $$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$ has three positive zeros $a,b,c$ :-)