- #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
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.
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.