To prove the "##m^{\text{th}}## Powers Theorem"

[brotherbobby]

Homework Statement

My textbook has listed the following theorem, calling it the "$m^{\text{th}}$ Powers Theorem".

If $a_1, a_2, \dots, a_n$ be a set of positive numbers not all equal, then
1. $$\boxed{\frac{\left( \sum_{i=1}^{n} a_i^m \right)}{n} < \left(\frac{\sum_{i=1}^{n}a_i}{n}\right)^m}$$, when $0<m<1$
2. $$\boxed{\frac{\left( \sum_{i=1}^{n} a_i^m \right)}{n} > \left(\frac{\sum_{i=1}^{n}a_i}{n}\right)^m}$$ when $m\in \mathbb{R}-(0,1)$

Relevant Equations

I am not sure what the relevant equations to prove the above identities are

Statement : Let me copy and paste the statement as it appears in the text on the right.

Attempt : I could attempt nothing to prove the identity. The best I could do was to verify it for a given value of the $a's, m, n$. I am not even sure what this identity is called but I will take the author's word for it - "The mth Powers Theorem".

Verify :

(1) Let some $m=0.5 (<1)$, $n=3$ and $a_i's = \{2,3,4\}$. Then the L.H.S. = $\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{3}$ = 1.72. The R.H.S. = $\left( \frac {2+3+4}{3} \right)^{0.5}$ = 1.73. Hence we see that L.H.S < R.H.S.

(2) Let some $m=2 (>1)$, $n=3$ and $a_i's = \{2,3,4\}$. Then the L.H.S. =$\frac{2^2+3^2+4^2}{3}$ = 9.67. The R.H.S. = $\left( \frac {2+3+4}{3} \right)^2$ = 9. Hence we see that L.H.S > R.H.S.

So the theorem is probably true but we can't be sure.

Moreover, let's see on the last line for the stipulation for $m$ which I copy and paste to the right.
This implies that $m<0$, say $m= - 0.5$. I haven't verified this case but let's assume that the theorem holds for it.

Request : A hint or help to help prove these two identities would be welcome.

 

[PeroK]

What about induction on $n$?

 

[brotherbobby]

Thank you. Let me try.

 

[martinbn]

The idea is that $x^m$ is convex and concave for the corresponding values of $m$. If you draw the graph you can see why it holds. These type of inequalities go by the name Jensen's inequality. The wiki article on it is not bad.