Problem: Suppose $a_1 , a_2 , \dots , a_p$ be fixed positive numbers. Consider sequences $s_n = \displaystyle\frac{{a_1}^n + {a_2}^n + \dots + {a_p}^n}{p}$ and $x_n = \sqrt[n]{s_n}$, $n \in \mathbb{N}$. Prove that ${x_n}$ be a monotone increasing sequence.
My attempt: Consider the following sequence $\displaystyle\left\{ \frac{s_n}{s_{n-1}} \displaystyle\right\}$ we have $\displaystyle\left\{ \frac{s_n}{s_{n-1}} \displaystyle\right\} = \frac{{a_1}^n + {a_2}^n + \dots + {a_p}^n}{p} . \frac{p}{{a_1}^{n-1} + {a_2}^{n-1} + \dots + {a_p}^{n-1}} = \frac{{a_1}^n + {a_2}^n + \dots + {a_p}^n}{{a_1}^{n-1} + {a_2}^{n-1} + \dots + {a_p}^{n-1}} > 1$.
Hence $\{s_n\}$ is a monotone incresing sequence. Then sequence $x_n = \sqrt[n]{s_n} = \displaystyle\sqrt[n]{\frac{{a_1}^n + {a_2}^n + \dots + {a_p}^n}{p}}$ is also a monotone incresing sequence.
Is this proof true? Thank all!
We need to prove that $$\left(\frac{\sum\limits_{k=1}^{p}x_k^{n+1}}{p}\right)^{\frac{1}{n+1}}\geq\left(\frac{\sum\limits_{k=1}^{p}x_k^n}{p}\right)^{\frac{1}{n}}$$ or $$p\left(\sum\limits_{k=1}^{p}x_k^{n+1}\right)^{n}\geq\left(\sum\limits_{k=1}^{p}x_k^n\right)^{n+1},$$ which is true by Holder: $$p\left(\sum\limits_{k=1}^{p}x_k^{n+1}\right)^{n}=\sum_{k=1}^p1\left(\sum\limits_{k=1}^{p}x_k^{n+1}\right)^{n}\geq$$ $$\geq\left(\sum_{k=1}^p\sqrt[n+1]{1\cdot \left(x_k^{n+1}\right)^n}\right)^{n+1}=\left(\sum\limits_{k=1}^{p}x_k^n\right)^{n+1}.$$ Done!