- #1

brotherbobby

- 700

- 163

- 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

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

**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

(1)

**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.

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.