For all $n \in \mathbb{N}^+$ and $(a_1, \ldots, a_n) \in [\mathbb{R}^+]^n$, we have $$\sum_{i=1}^n a^n_i \ge n \prod_{i=1}^n a_i$$
I posted a question about AM-GM inequality in $3$ variables here. After that, I found that I can utilize the same method to generalize this inequality to arbitrary $n$ variables.
Could you please verify if my attempt contains logical gaps/errors?
My attempt:
We prove this statement by induction on $n$. Clearly, it holds for $n = 1$. Let it hold for $n$. We have
$$\begin{aligned} &\left ( \sum_{i=1}^{n+1} a^{n+1}_i \right ) + (n-1)\prod_{i=1}^{n+1} a_i \\ = & \left ( \sum_{i=1}^{n} a^{n+1}_i \right ) + \left ( a_{n+1}^{n+1} + \underbrace{\prod_{i=1}^{n+1} a_i + \ldots \prod_{i=1}^{n+1} a_i}_{(n-1) \text{ times}} \right ) \\ \ge & n \sqrt[n]{\prod_{i=1}^n a_i^{n+1}} + n \sqrt[n]{a_{n+1}^{n+1} \left (\prod_{i=1}^{n+1} a_i \right )^{n-1}} \quad \text{ by inductive hypothesis}\\ = & n \sqrt[n]{\prod_{i=1}^n a_i^{n+1}} + n \sqrt[n]{a_{n+1}^{n+1} \prod_{i=1}^{n+1} a_i^{n-1}} \\ = &n \left (2 \sqrt[2]{\sqrt[n]{\left ( \prod_{i=1}^n a_i^{n+1} \right ) a_{n+1}^{n+1} \prod_{i=1}^{n+1} a_i^{n-1} }} \right )\\ = & 2n \sqrt[2]{\sqrt[n]{\prod_{i=1}^{n+1} a_i^{2n}}} \\ = & 2n \prod_{i=1}^{n+1} a_i \end{aligned}$$
It follows that $$\sum_{i=1}^{n+1} a^{n+1}_i \ge [2n - (n-1)] \prod_{i=1}^{n+1} a_i = (n+1) \prod_{i=1}^{n+1} a_i$$
Then the statement holds for $n+1$. This completes the proof.
