I don't see how Cauchy's proof of AM $\ge$ GM holds for all cases? I am reading Maxima and Minima Without Calculus by Ivan Niven and on pages $24-26$ he gives Cauchy's proof for the $AM-GM$ . The general idea of the proof is that $P_{n}$ is the proposition $$(a_{1}+a_{2}+\cdot \cdot \cdot a_{n}) \ge n(a_{1} a_{2} \cdot \cdot \cdot  a_{n})^{1/n}$$
The proof proves that if $P_{n}$ holds, then $P_{n-1}$ and $P_{2n}$ also hold. I have a problem with the part that shows that $P_{n-1}$ holds. It goes something like this:
$g$ is the geometric mean of $a_{1} , a_{2} , \cdot \cdot \cdot a_{n-1}$ so that $g=(a_{1} , a_{2} , \cdot \cdot \cdot a_{n-1})^{1/n-1}$
Replace $a_{n}$ with $g$:
$( a_{1}+ a_{2}+ \cdot \cdot \cdot +a_{n-1}+ g) \ge n ( a_{1} a_{2} \cdot \cdot \cdot  g)^{1/n}$ 
The $RHS$ becomes $n(g^{n-1} g)^{1/n}=ng$
So we have $( a_{1}+ a_{2}+ \cdot \cdot \cdot +a_{n-1}+ g) \ge ng$ , or $( a_{1}+ a_{2}+ \cdot \cdot \cdot +a_{n-1}) \ge (n-1) g$ , therefore if $P_{n}$ holds, then $P_{n-1}$ does also.
I understand the logic of the proof, but the problem I have is when he replaced $a_{n}$ with $g$ . Obviously, the last term is not necessarily equal to the the geometric mean of the rest of the terms. So how does this proof work for all cases? Thanks.

We wish to show that the case of the inequality with $n$ non-negative numbers proves the case with $n-1$ numbers. Take any $n-1$ non-negative numbers. We may put any non-negative number for $a_n$ since we already know it to be true in the case of $n$ numbers. We simply choose $a_n=g$ where $g$ is as you defined it.
