The A.M.-G.M. is sometimes stated as follows.
Let $a$ and $b$ are positive numbers such that $b < a$. Let $a_1 = a$ and $b_1 = b$. Define the sequences $\{a_n\}$ and $\{b_n\}$ recursively for all natural numbers $n\geq 2$ by $$a_{n+1} = \frac{a_n + b_n}{2} \hspace{5mm} \text{and} \hspace{5mm} b_{n+1} = \sqrt{a_n b_n}$$ Then we have the inequality for all $n \in \mathbb{N}$ $$b_n \leq b_{n+1} \leq a_{n+1} \leq a_n$$
Why do we require b < a? If $a < b$, I observed that for $n\geq 2$, the above inequality must be rewritten as $$a < b_n \leq b_{n+1} \leq a_{n+1} \leq a_n < b$$ I couldn't find anything wrong with the assumption $a < b$.
If $a<b$ then
$$a_2=\frac{a+b}{2}$$ and $$b_2=\sqrt{ab}$$
If $a>b$ then
$$a_2=\frac{b+a}{2}$$ and $$b_2=\sqrt{ba}$$
So, for $n\ge 2$, the sequence terms do not depend on initial conditions.
$(a_n)$ and $(b_n)$ are adjacent and the common limit do not depend on $a>b$ or $a<b$.