product of $f(x)f(y)f(z)\geq1$ for $f(x)=ax^2+bx+c$ with $xyz=1$ and $a+b+c=1$.

Question

Let $a,b,c,x,y,z$ be positive reals with $xyz=1$ and $a+b+c=1$.

Let $f(x)=ax^2+bx+c$. Prove $f(x)f(y)f(z)\geq 1$.

Im stuck, I tried a bunch of things but the best ive managed to obtain is:

$f(x)f(y)f(z)\geq 27\sqrt[9]{abc}$ which is useless.

Answer 1

By Holder: $$(ax^2+bx+c)(ay^2+by+c)(az^2+bz+c)\geq\left(a\sqrt[3]{x^2y^2z^2}+b\sqrt[3]{xyz}+c\right)^3=1.$$