-4
$\begingroup$

I can't find any proof of this inequality $\displaystyle \frac{a_1^m+a_2^m+\cdots+a_n^m}{n}\ge \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^m$ .

Please any help is appreciated.

  • 1
    Check the book "The Cauchy Schwarz Master Class" by J. Michael Steele; it's an exercise on page $36$. The book has solutions to exercises. – 2017-02-06
  • 1
    [Inequality between any two power means](https://en.wikipedia.org/wiki/Generalized_mean#Inequality_between_any_two_power_means) – 2017-02-06
  • 1
    I think you can use [Holder's inequality](http://mathworld.wolfram.com/HoeldersInequalities.html). – 2017-02-06
  • 1
    The author offers a beautiful argument; he uses a technique called "backwards induction". – 2017-02-06
  • 0
    I think your original inequality was incorrect. I changed it. – 2017-02-06
  • 0
    @S.C.B. Indeed, because $a_1=a_2=\cdots = a_n$ gives equality. – 2017-02-06

3 Answers 3

1

I'm assuming $a_1,a_2,\ldots,a_n>0$ and $m,n$ are positive integers.

By Hölder’s inequality: $$\left(\underbrace{1+1+\cdots+1}_{n\text{ times}}\right)^{m-1}\times $$

$$\times \left(a_1^m+a_2^m+\cdots+a_n^m\right)$$

$$\ge \left(a_1+a_2+\cdots+a_n\right)^m$$

Equality holds at least when $a_1=a_2=\cdots=a_n$.

1

Your inequality is wrong! Try $m=n=3$, $a_2=a_3=...=a_m=0$ and $a_1=-1$.

We have $-\frac{1}{3}\geq-\frac{1}{27}$, which is wrong.

  • 0
    Every entry is positive , -1 won't work – 2017-02-06
  • 1
    @AdityaNarayanSharma that was a call to update the question. – 2017-02-06
0

Using one of the properties of convex functions, more precisely: $$f\left( \frac{1}{n}\sum_{i=1}^{n}x_n \right)\leq \frac{1}{n}\sum_{i=1}^{n}f(x_n)$$ It's simple to test $f(x)=x^m$ for convexity.

On top of this, if $m$ is even $f''(x)\geq 0$ or the function is convex everywhere and $a_i$ can be negative and positive. For odd $m$, the inequality works for positive $a_i$ only.

I can suggest this book for further reading, it has a chapter dedicated to convex functions. If you google for it, you will probably find a PDF freely available. You will also find the $n=2$ case there, as an exercise.