AM-GM inequality using Lagrange multipliers- how is existence of extrema really confirmed? 
Prove the following inequality $$({x_1x_2,\ldots,x_n})^{1\over n}\le {1\over n}({x_1+\cdots+x_n})$$ using Lagrange multipliers.

While I do understand how to find the extremum points, I don't quite understand what the grounds for the Theorem itself are. I mean, an extrema point, a local one, has to be declaired existent, and no other solutions have been clear to me in that matter.
Can I simply assume there is and that $f(x)=\lambda g(x)$, with, say, $f(x)={1\over n}(x_1+\cdots+x_n)$ and $g(x)=x_1x_2\cdot x_n-c$ (assuming that for given $x_i\ge 0$, $x_1\cdot x_n=c$ for some $c$).
Something in the order of proccesses seems unsupported or unfounded. I would appreciate it if you could make things clearer for me. 
 A: Considering $f(x) = x_1  \cdots x_n $ and $\varphi (x) = x_1 + \ldots + x_n$. Fix $s > 0$ and we want to find the critical points (at first) of $f|_{M}$, where $M = \varphi^{-1}(s)$.
Well, $\mathrm {grad} \varphi (x) = (1,\ldots, 1) $ and $\mathrm {grad} f(x) = (a_1, \ldots, a_n)$, with $a_j = \displaystyle \prod_{j \neq i} x_j$. 
Now, $x \in M$ is a critical point, iff, for some $\lambda$, we have $$\prod_{j\neq i} x_j = \lambda, \,\,\,\ i  = 1, \ldots, n$$
Diving the $i$-th of these equations by the $k-$th gives us $\frac{x_j}{x_ i} = 1 $. Hence the only critical point of $f|_M$ is the one with the same coordinate, that is , $ p = (s/n , \ldots, s/n)$. 
Notice that $f$ is continuous on the compact $\overline M$, then by Weierstrass Theorem it has a maximum and it cannot  be on $\overline M - M$, because $$x \in \overline M - M  \implies x_1 \cdot x_2 \cdot \cdots x_n = 0 $$
(it helps if you sketch something in $ \mathbb R^3$). Thus we may say that the maximum point is, indeed, in $M$, and it is $f(p) = (s/n)^n$. 
Finally if $x_1, \ldots, x_n$ are positive then $$x_1\cdot x_2 \cdots x_n \leq \left(\frac{s}{n}\right)^n = \left(\frac{x_1 + x_2+ \ldots + x_n}{n}\right)^n$$
