Min k_n satisfying an Inequality

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Min k_n satisfying an Inequality

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Nov 11, 2010, 6:41 AM • 2 Y

Y by Adventure10, Mango247

For $0<x_1<1,\ 0<x_2<2,\ \cdots ,\ 0<x_n<n$ ,

Determine the minimum value of the constant $k_n$ such that :

$x_1x_2\cdots x_n+k_n\geq n!\left(\frac {x_1}{1}+\frac{x_2}{2}+\cdots +\frac{x_n}{n}\right).$

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kuing

#2 h Nov 11, 2010, 7:10 AM • 1 Y

Y by Adventure10

kunny wrote:

For $0<x_1<1,\ 0<x_2<2,\ \cdots ,\ 0<x_n<n$ ,

Determine the minimum value of the constant $k_n$ such that :

$x_1x_2\cdots x_n+k_n\geq n!\left(\frac {x_1}{1}+\frac{x_2}{2}+\cdots +\frac{x_n}{n}\right).$

for each $x_k,(1\le k \le n)$ , we have

$\begin{aligned} f(x_k)&=n!\left(\frac {x_1}{1}+\frac{x_2}{2}+\cdots +\frac{x_n}{n}\right)-x_1x_2\cdots x_n\\ &=\left(\frac{n!}{k}-\frac{\prod_{i=1}^n{x_i}}{x_k}\right)x_k + C \end{aligned}$

where $C$ is not related to $x_k$ .

from the conditions, easy to know $\frac{n!}{k}>\frac{\prod_{i=1}^n{x_i}}{x_k}$ , so we see that
$n!\left(\frac {x_1}{1}+\frac{x_2}{2}+\cdots +\frac{x_n}{n}\right)-x_1x_2\cdots x_n$
is increasing of each variable. thus, we get
$\sup \left\{ n!\left(\frac {x_1}{1}+\frac{x_2}{2}+\cdots +\frac{x_n}{n}\right)-x_1x_2\cdots x_n \right\}=n\cdot n!-n!=(n-1)\cdot n!$ ,
that's the min $k_n$ .

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Kimpul

#3 h Nov 11, 2010, 7:25 AM • 1 Y

Y by Adventure10

alternatively, define $y_i=\frac{x_i}{i}$ for $i=1,2,\ldots,n$ . so $0<y_i<1$ .

the following inequality can be easily proved using induction:

$y_1y_2\cdots y_n + n-1\geq y_1+y_2+\cdots +y_n$ .

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