Anika Boyd

2022-03-30

$f\left(x\right)=a{x}^{2}+bx+c$ are positive, and $a+b+c=1$
Prove $f\left({x}_{1}\right)f\left({x}_{2}\right)\cdots f\left({x}_{n}\right)\ge 1$ is true for all positive x, satisying ${x}_{1}{x}_{2}\cdots {x}_{n}=1$

Ashley Olson

Step 1
It's wrong.
For $n=2$ the following is true.
$\underset{x>0,y>0,xy=1}{min}\left(5{x}^{2}-14x+10\right)\left(5{y}^{2}-14y+10\right)=\frac{1}{2}.$
The equality occurs for example, for
$\left(x,y\right)=\left(\frac{21-\sqrt{41}}{20},\frac{21+\sqrt{41}}{20}\right).$
By the way, $a\ge 0$ and $c\ge 0$ by the given.
Also, for $b\ge 0$ we can use Holder here:
$\prod _{k=1}^{n}\left(a{x}_{k}^{2}+b{x}_{k}+c\right)\ge \left(a\sqrt[n]{\prod _{k=1}^{n}{x}_{k}^{2}}+b\sqrt[n]{\prod _{k=1}^{n}{x}_{k}}+c{\right)}^{n}$
$={\left(a+b+c\right)}^{n}=1$

Jared Kemp

Step 1
(Compare A Quadratic With Positive Coefficients or Inequalities with polynomials on AoPS). Any polynomial f with real positive coefficients satisfies
$f\left({x}_{1}\right)\cdots f\left({x}_{n}\right)\ge f\left(\sqrt[n]{{x}_{1}\cdots {x}_{n}}{\right)}^{n}$
for positive numbers ${x}_{1},\dots ,{x}_{n}$. That is a consequence of the generalized Holder inequality, alternatively one can prove that $y\to \mathrm{log}\left(f\left({e}^{y}\right)\right)$ is convex and apply Jensen's inequality.
In our case is ${x}_{1}\cdots {x}_{n}=1$ and ${x}_{1}\cdots {x}_{n}=1$, so that ${x}_{1}\cdots {x}_{n}=1$.

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