To prove the AM GM inequality, one would firstly prove the $2^k$ case such that $ a_1 a_2 a_3...a_{2^k} \leq \frac{a_1+ a_2+ a_3...+a_{2^k} }{2^k}$ given by induction. To prove the odd cases, the sequence would be padded by the average, $A$
$A \equiv \frac{a_1+ a_2+ a_3...+a_{n} }{n}$,
such that there are enough copies of the average $A$ onto the $2^k$ case of the AM GM inequality, as shown.
$ [a_1 a_2 a_3...a_{n} × A^{2^k-n}]^\frac{1}{2^k} \leq \frac{a_1+ a_2+ a_3...+a_{n} +(2^k-n)A }{2^k} = \frac{2^k ×A}{2^k} = A$
Resolving the $LHS$ and $RHS$ for $A$ provides the AM GM inequality.
My question pertains to the value of $A$ itself. Wouldn't the value be arbitrary in the sense that changing the value of A would yield an arbitrary bound? Why is the average of the values used here instead (in the sense of how would one know beforehand to purposely select its average)?
Instead of thinking of it as padding enough copies of $A,$ I prefer to think of the second step of Cauchy induction as proving $P(n-1)$ using $P(n)$ where $P$ is the proposition. Then it becomes a matter of using the decomposition $$\frac{a_1 +a_2+\cdots +a_{n-1}}{n-1} = \frac{1}{n}\left(a_1+a_2+\cdots+a_{n-1}+\frac{a_1+a_2+\cdots+a_{n-1}}{n-1}\right).$$ You can use $P(n)$ on the right side and then some straightforward manipulations give you $P(n-1).$ Does that make the argument more natural for you?