I saw this question in my book
Let $a_1 , a_2 , \dots,a_m$ be fixed positive numbers and $S_n =\frac{\sum \limits_{k=1}^ m a_k^{n}}{m}$ Prove that $\sqrt[n] {S_n} $ is monotone increasing sequence
Hint first establish monotonicity of the sequence $\frac{S_n}{S_{n-1}}$
I couldn't prove that hint or know how to use it solve the question But it is very easy to prove The question using Induction and Cauchy's inequality
Proving $\sqrt[n] {S_n}$ is monotone increasing sequence is the same as proving $$S_n^{n+1}\le S_{n+1}^n$$
$$\frac{\left(\sum \limits_{k=1}^ m a_k^{n}\right)^{n+1} }{m^{n+1}} \le \frac{\left(\sum \limits_{k=1}^ m a_k^{n+1}\right)^{n} }{m^{n}} $$
$${\left(\sum \limits_{k=1}^ m a_k^{n}\right)^{n+1} } \le m{\left(\sum \limits_{k=1}^ m a_k^{n+1}\right)^{n} }$$
Let $P_n =m S_n$.
By Cauchy's inequality $P_1 ^2 \le m P_2$
Assume $(P_n)^{n+1}\le m P_{n+1}^n $, so $P_n \le m^{\frac{1}{n+1}} P_{n+1}^{\frac{n}{n+1}}$
since $n+1 = \frac{n+2+n}{2}$
by Cauchy's inequality
$$P_{n+1}^2 \le P_{n+2} P_n $$
$$P_{n+1}^2 \le P_{n+2} \times m^{\frac{1}{n+1}} \times P_{n+1}^{\frac{n}{n+1}}$$
$$P_{n+1}^{n+2} \le m P_{n+2} ^{n+1}$$
This seem to be easier proof that what the book wanted
As I mentioned before I was not able to prove the hit but it is obvious when all of $a_k$ is less than $1$ or all of them are greater than 1
I want to ask for a prove for the hint and how to use it to prove the question and a third question is : is my proof correct ?
Best Answer
For monotonicity,we just need to prove$\frac{\frac{S_{n+1}}{S_{n}}}{\frac{S_{n}}{S_{n-1}}}={\frac{S_{n+1}S_{n-1}}{S_{n}^2}}\geq1$,which is trivial by Cauchy.
For the whole proof,we just need to see$\sqrt[n]{S_n}=\sqrt[n]{\frac{S_{n}}{S_{n-1}}\frac{S_{n-1}}{S_{n-2}}...\frac{S_{1}}{S_{0}}}$which is a geometric mean that is clearly monotone.
I didn't find any error in your proof.I think it's ok;D