$f(x) = ax^2+bx+c$ are positive, and $a+b+c=1$. Prove $f(x_1)f(x_2)...f(x_n)\geq1$ is true for all positive $x$, satisfying $x_1x_2...x_n=1$

Question

I was working through some past paper questions when this question came up. The question is as follows.

$f(x) = ax^2+bx+c$ are positive, and $a+b+c=1$. Prove that the inequality $f(x_1)f(x_2)...f(x_n)\geq1$ holds for all positive numbers $x_1,x_2...x_n$, satisfying $x_1x_2...x_n=1$

I tried proving that $f(x)\geq x$ for all positive $x$, which I was able to easily prove for all $x\geq 1$. However, I found it hard to prove that $f(x) \geq x$ for $0<x<1$. Can anyone give some thoughts on how to prove this question?

Answer 1

(Compare A Quadratic With Positive Coefficients or Inequalities with polynomials on AoPS). Any polynomial $f$ with real positive coefficients satisfies $$ f(x_1)\cdots f(x_n) \ge f(\sqrt[n]{x_1 \cdots x_n})^n $$ for positive numbers $x_1, \ldots, x_n$. That is a consequence of the generalized Hölder inequality, alternatively one can prove that $y \to \log(f(e^y))$ is convex and apply Jensen's inequality.

In our case is $x_1 \cdots x_n = 1$ and $f(1) = 1$, so that $f(x_1)\cdots f(x_n) \ge 1$.

Answer 2

It's wrong.

For $n=2$ the following is true. $$\min_{x>0,y>0,xy=1}(5x^2-14x+10)(5y^2-14y+10)=\frac{1}{2}.$$ The equality occurs for example, for $$(x,y)=\left(\frac{21-\sqrt{41}}{20},\frac{21+\sqrt{41}}{20}\right).$$

By the way, $a\geq0$ and $c\geq0$ by the given.

Also, for $b\geq0$ we can use Holder here: $$\prod_{k=1}^n(ax_k^2+bx_k+c)\geq\left(a\sqrt[n]{\prod_{k=1}^nx_k^2}+b\sqrt[n]{\prod_{k=1}^nx_k}+c\right)^n=(a+b+c)^n=1$$