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.
 A: The last term is not necessarily equal to the geometric mean of the rest that's right , but we are free to choose any term $a_i$ as long as $a_i >0$. Hence for this proof we set $g:=a_n$ in order to peform induction. Then we start our induction with $n=2$ wich folows from the well known ineuqality $(x-y)^2\geq 0 \Leftrightarrow x^2+y^2 \geq 2xy$. So for $n=2$ the inequality holds. 
Then your book proofs: If it is valid for $n$ it is valid for $n-1$ hence the $AM-GM$ inequality is valid for $n=1$. Then it proofs ( wich you didn't post) that if it is valid for $n$ it is valid for $2n$ .
So it is valid for $n:=2$ , $n-1=1$ , $2n=4$. Now set $n:=4$ and it will be valid for $n-1=3$. Set $n:=3$ and it will be valid for $2n=6$ and so on.. You can see that we will cover all natural numbers like that.
A: 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.
