Problematic step
$$P(n) \implies p(n-1)$$ Here we assume P is true $\forall n : 0\le n \le n + 1$
So,
$${a_1 + a_2 + \cdots+ a_n \over n} = \sqrt[n]{a_1 a_2 \cdots a_n}$$
let $$a_n ={ a_1 + a_2 + \cdots+ a_{n-1} \over n-1}$$
then we do little bit of algebraic manipulation to get the AM-GM inequality.
Doubt
When taking $a_n$ specifically as ${ a_1 + a_2 + \cdots+ a_{n-1} \over n-1}$, aren't we saying that AM-GM inequality is only true when any one term is mean of all other terms ([1]) ? aren't we suppose to prove the inequity with some general $a_n$ not a specific case.
counter example- [1] $$ {1+2+7\over 3} = 10/3 \ge \sqrt[3]{1*2*7} = \sqrt[3]{14}$$ $$7 \ne {1+2\over 2}, 2 \ne {1+7\over 2},1 \ne {7+2\over 2}$$
A bit of clarification, if i sound vague
In induction proof of AM-GM by taking a specific $a_n$, how we are concluding hat it hold for all $a_n \in\mathbb{R}$
I am not saying that i broke maths or whatever, just want a bit of clarification.Thanks $\ddot \smile$