<s>[INST] product of $f(x)f(y)f(z)\geq1$ for $f(x)=ax^2+bx+c$ with $xyz=1$ and $a+b+c=1$. 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.
 [/INST] By Holder:
$$(ax^2+bx+c)(ay^2+by+c)(az^2+bz+c)\geq\left(a\sqrt[3]{x^2y^2z^2}+b\sqrt[3]{xyz}+c\right)^3=1.$$
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