Simple inequality

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Simple inequality

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sqing

#1 h Jul 21, 2018, 1:45 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c\ge 0 .$ For the integer $n\ge3$ ,prove that $a^{n+1}+b^{n+1}+c^{n+1}+1\geq 2(a^n+b^n+c^n)-(a^{n-1}+b^{n-1}+c^{n-1})+1\geq a^n+b^n+c^n+abc.$ here

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#2 h Jul 21, 2018, 1:50 PM • 2 Y

Y by Adventure10, Mango247

I think $2(a^n+b^n+c^n)-(a^{n-1}+b^{n-1}+c^{n-1})+1\geq a^n+b^n+c^n+abc$ is wrong

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41157 posts

sqing

#3 h Jul 21, 2018, 2:15 PM • 2 Y

Y by Adventure10, Mango247

delfinom wrote:

I think $2(a^n+b^n+c^n)-(a^{n-1}+b^{n-1}+c^{n-1})+1\geq a^n+b^n+c^n+abc$ is wrong

Are you sure?

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#4 h Jul 21, 2018, 2:24 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

delfinom wrote:

I think $2(a^n+b^n+c^n)-(a^{n-1}+b^{n-1}+c^{n-1})+1\geq a^n+b^n+c^n+abc$ is wrong

Are you sure?

Sorry it's true, here is my solution $\sum \frac{(n-1)a^n+1}{n}+\frac{a^n+b^n+c^n+n-3}{n}\geq (\sum a^{n-1})+abc$

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41157 posts

sqing

#5 h Jul 21, 2018, 2:38 PM • 2 Y

Y by Adventure10, Mango247

delfinom wrote:

sqing wrote:

delfinom wrote:

I think $2(a^n+b^n+c^n)-(a^{n-1}+b^{n-1}+c^{n-1})+1\geq a^n+b^n+c^n+abc$ is wrong

Are you sure?

Sorry it's true, here is my solution $\sum \frac{(n-1)a^n+1}{n}+\frac{a^n+b^n+c^n+n-3}{n}\geq (\sum a^{n-1})+abc$

Yeah.

Let $a,b,c\ge 0 .$ For the integer $n\ge3$ ,we have $a^n+b^n+c^n+1\geq a^{n-1}+b^{n-1}+c^{n-1}+abc.$

This post has been edited 1 time. Last edited by sqing, Jul 21, 2018, 2:44 PM

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xzlbq

#7 h Jul 21, 2018, 4:12 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c,d>0,n\geq 4$ ,prove that

${a}^{n}+{b}^{n}+{c}^{n}+{d}^{n}+1-{a}^{n-1}-{b}^{n-1}-{c}^{n-1}-{d}^{n -1}-abcd\geq 0$

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15849 posts

xzlbq

#8 h Jul 21, 2018, 4:14 PM • 2 Y

Y by Adventure10, Mango247

Let $x_i>0,i=1,...,n,n\geq 3,t\geq n$ ,prove that

$\sum_{i=1}^n{x_i^t}+1\geq \sum_{i=1}^n{x_i^{t-1}}+\prod_{i=1}^n{x_i}$

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41157 posts

sqing

#9 h Jul 22, 2018, 12:06 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

Let $x_i>0,i=1,...,n,n\geq 3,t\geq n$ ,prove that

$\sum_{i=1}^n{x_i^t}+1\geq \sum_{i=1}^n{x_i^{t-1}}+\prod_{i=1}^n{x_i}$

Simple inequality，too.

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MYL

#10 h Jul 22, 2018, 7:42 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

Let $x_i>0,i=1,...,n,n\geq 3,t\geq n$ ,prove that

$\sum_{i=1}^n{x_i^t}+1\geq \sum_{i=1}^n{x_i^{t-1}}+\prod_{i=1}^n{x_i}$

$\sum_{i=1}^{n} {x_i^t} + 1 = \sum_{i=1}^{n} {\left( \frac{t-1}{t}x_i^t + \frac{1}{t} \right)} + \frac{1}{t} \left( (t-n) + \sum_{i=1}^{n} {x_i^t} \right) \ge \sum_{i=1}^n {x_i^{t-1}} + \prod_{i=1}^n{x_i}$

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41157 posts

sqing

#11 h Jul 22, 2018, 7:47 AM • 2 Y

Y by Adventure10, Mango247

MYL wrote:

xzlbq wrote:

Let $x_i>0,i=1,...,n,n\geq 3,t\geq n$ ,prove that

$\sum_{i=1}^n{x_i^t}+1\geq \sum_{i=1}^n{x_i^{t-1}}+\prod_{i=1}^n{x_i}$

$\sum_{i=1}^{n} {x_i^t} + 1 = \sum_{i=1}^{n} {\left( \frac{t-1}{t}x_i^t + \frac{1}{t} \right)} + \frac{1}{t} \left( (t-n) + \sum_{i=1}^{n} {x_i^t} \right) \ge \sum_{i=1}^n {x_i^{t-1}} + \prod_{i=1}^n{x_i}$

Yeah.
Very simple.

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