Anika Boyd

2022-03-30

Prove

Ashley Olson

Beginner2022-03-31Added 12 answers

Step 1

It's wrong.

For $n=2$ the following is true.

$\underset{x>0,y>0,xy=1}{min}(5{x}^{2}-14x+10)(5{y}^{2}-14y+10)=\frac{1}{2}.$

The equality occurs for example, for

$(x,y)=(\frac{21-\sqrt{41}}{20},\frac{21+\sqrt{41}}{20}).$

By the way, $a\ge 0$ and $c\ge 0$ by the given.

Also, for $b\ge 0$ we can use Holder here:

$\prod _{k=1}^{n}(a{x}_{k}^{2}+b{x}_{k}+c)\ge (a\sqrt[n]{\prod _{k=1}^{n}{x}_{k}^{2}}+b\sqrt[n]{\prod _{k=1}^{n}{x}_{k}}+c{)}^{n}$

$={(a+b+c)}^{n}=1$

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

$f\left({x}_{1}\right)\cdots f\left({x}_{n}\right)\ge f(\sqrt[n]{{x}_{1}\cdots {x}_{n}}{)}^{n}$

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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?

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