Use AM-GM to show $\frac{a_1^{n+1}+\dots+\ a_n^{n+1}+1}{n+1}\ge\frac{a_1^n+\dots+a_n^n}{n}$

Question

(Only AM-GM is allowed) Show $$\frac{a_1^{n+1}+\dots+\ a_n^{n+1}+1}{n+1}\ge\frac{a_1^n+\dots+a_n^n}{n}$$ where $\{a_i\}_{i=1}^n$ is a positive sequence with sum $n$

It is easy to show it with Chebyshev (or rearrangement) inequality, I'm just wondering, how to apply AM-GM here

Answer 1

By AM-GM, $na_k^{n+1}+1\geq (n+1)a_k^n ,$ or $a_k^{n+1}+\frac{1}{n}\geq\frac{n+1}{n}a_k^n$. So $$\dfrac{\sum_{k=1}^n (a_k^{n+1}+\frac{1}{n})}{n+1}\geq\dfrac{1}{n+1}\sum_{k=1}^n \frac{n+1}{n}a_k^n = \dfrac{a_1^n+a_2^n+...+a_n^n}{n}$$