Anika Boyd

## Answered question

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 ${x}_{1}{x}_{2}\cdots {x}_{n}=1$

### Answer & Explanation

Ashley Olson

Beginner2022-03-31Added 12 answers

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

Beginner2022-04-01Added 14 answers

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 ${x}_{1}\cdots {x}_{n}=1$.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?