The Arithmetic-Geometric Mean Inequality
Theorem 1. Let $a_1, ..., a_n > 0$. Then $$\frac{a_1 + \cdots + a_n}{n} \geq (a_1 \cdots a_n)^{\frac{1}{n}},$$ with the equality holds $\iff$ $a_j = a_1, \forall j, 2 \leq j \leq n$.
Proof. Let $S$ be the set of all natural numbers which satisfy the given inequality. Cleary $1 \in S$.
Firstly, we claim that $\{2^k \, | \, k \in \mathbb{N} \} \subseteq S$.
$(\because)$ We may use the principle of the mathematical induction. Obviously, $2^1 \in S$ which is the case for $k = 1$. $(2 \sqrt{a_1a_2}$ $\leq a_1 + a_2$ $\iff 4a_1a_2$ $\leq a_1^2 + 2a_1a_2 + a_2^2$ $\iff$ $(a_1 - a_2)^2 \geq 0)$
Suppose that $2^k \in S$ for some $k \in \mathbb{N}$. We want to show that $2^{k+1} \in S$ which means that $$(a_1 \cdots a_{2^{k+1}})^{\frac{1}{2^{k+1}}} \leq \frac{a_1 + \cdots + a_{2^{k+1}}}{2^{k+1}}.$$ Then we have $$\sqrt{(a_1 \cdots a_{2^k})^{\frac{1}{2^k}} (a_{2^k+1} \cdots a_{2^{k+1}})^{\frac{1}{2^k}}} \leq \frac{(a_1 \cdots a_{2^k})^{\frac{1}{2^k}} + (a_{2^k+1} \cdots a_{2^{k+1}})^{\frac{1}{2^k}}}{2} \\ \leq \frac{\frac{a_1 + \cdots + a_{2^k}}{2^k} + \frac{a_{2^k+1} + \cdots + a_{2^{k+1}}}{2^k}}{2} = \frac{a_1 + \cdots + a_{2^{k+1}}}{2^{k+1}},$$ which shows that $2^{k+1} \in S$.
Secondly, we claim that $n \in S \Longrightarrow n-1 \in S (n > 1)$.
$(\because)$ Note that $$\frac{a_1 + \cdots a_{n-1}}{n-1} = \frac{(1 + \frac{1}{n-1})(a_1 + \cdots + a_{n-1})}{n} = \frac{a_1 + \cdots + a_{n-1} + \frac{a_1 + \cdots a_{n-1}}{n-1}}{n}.$$ Then we have $$\frac{a_1 + \cdots a_{n-1}}{n-1} = \frac{a_1 + \cdots + a_{n-1} + \frac{a_1 + \cdots a_{n-1}}{n-1}}{n} \geq \left( a_1 \cdots a_{n-1} \frac{a_1 + \cdots + a_{n-1}}{n-1} \right)^{\frac{1}{n}} \\ = (a_1 \cdots a_{n-1})^{\frac{1}{n}} \left( \frac{a_1 + \cdots + a_{n-1}}{n-1} \right)^{\frac{1}{n}} \\ \Longrightarrow \left( \frac{a_1 + \cdots a_{n-1}}{n-1} \right)^{1 - \frac{1}{n}} \geq (a_1 \cdots a_{n-1})^{\frac{1}{n}} \\ \Longrightarrow \frac{a_1 + \cdots a_{n-1}}{n-1} \geq (a_1 \cdots a_{n-1})^{\frac{1}{n-1}} \iff n-1 \in S.$$ Finally, we claim that $S = \mathbb{N}$.
$(\because)$ Let $n \in \mathbb{N}$. There exists $k \in \mathbb{N}$ such that $n < 2^k$. By our first claim, $2^k \in S$. Furthermore, by our second claim, $n \in S$. This means that $\mathbb{N} \subseteq S$. Thus $S = \mathbb{N}$. $\blacksquare$