Lagrange Multiplier In Proving AM-GM inequality

Question

I am dealing with a question in Stewart Calculus Book which states the following:

Using Lagrange multiplier, find the maximum of the function $$f(x_1,x_2,...,x_n)=(x_1x_2...x_n)^{1/n}$$subject to the constraint$$\sum_{i=1}^{n} x_i=c$$,where $x_i$'s are all positive, and use this to prove the AM-GM inequality. Actually in AM-GM inequality, it holds for all nonnegative numbers, but I think the author wants to get rid of the case where $x_i$'s are all zero, where Lagrange multiplier can no longer be applied.

And the result I obtained is: $f$ will have its maximum/minimum (the arithmetic mean), or even saddle point at $(\frac{c}{n},\frac{c}{n},...,\frac{c}{n})$, since Lagrange method gives no definiteness on the behaviour of the critical points found (whether max, min, or saddle).

And my question is: In this case, how do we show that the point I found gives the maximum value of the function? My teacher gave me an argument that said "the minimum value would be some place very near to $0$, so the arithmetic mean will not be the minimum value". But I think the argument is not persuasive enough to explain why the arithmetic mean is the maximum value. So, I hope that there will someone who are willing to give their opinions on this. Thanks.

Answer 1

Let $A$ be the set of $(x_1, x_2, ..., x_n)$ such that $$\sum x_i = c$$ and $x_i \geq 0$ for all $i$. On a point $x$ of the boundary of $A$, there is always a component $x_i$ of $x$ which is equal to zero, hence $f = 0$ on the boundary. As $f$ is continuous and $A$ is compact, $f$ has a global maximum on $A$. Denote by $x^\ast$ one such maximum. The fact that $f = 0$ on the boundary of $A$ implies that $x^\ast$ is in the interior of $A$. Then what you have done with Lagrange multipliers shows that $$x^\ast = (\frac{x_1}{c}, ..., \frac{x_n}{c})$$ as it is the only extremum of $f$ in the interior of $A$.

EDIT : Let me clarify my notations. Write $$\mathbb{R}^n_+ = \{ x = (x_1, ..., x_n), \ x_i \geq 0 \text{ for } i = 1, 2, ... n \}$$ $${\mathbb{R}^n_+}^\ast = \{ x = (x_1, ..., x_n), \ x_i > 0 \text{ for } i = 1, 2, ... n \}$$ and $$h(x) = \sum x_i - c.$$ Then $f$ has a maximum $x^\ast$ on the compact $$\{x \in \mathbb{R}^n_+, \ h(x) = 0 \}$$ and in fact $$ x^\ast \in \{x \in {\mathbb{R}^n_+}^\ast, \ h(x) = 0 \}$$ One can conclude using Lagange multipliers in this last set.