Proving AM-GM with the method of Lagrange multipliers

Question

In my calculus book, there is a question that basically says "use the method of Lagrange multipliers on $f(x,y,z)=xyz$ with constraint $g(x,y,z)=x+y+z=C$, $C$ being a constant, and use this to prove AM-GM for three variables." The next question asks to generalize this result.
It is very easy to see that the only possible location for an extremum of f along the constraint is when all three variables equal to each other, and that an other value of f along the constraint is less than this, but I couldn't think of a way to show that f actually has a maximum along the constraint short of independently proving AM-GM and applying it here.

Answer 1

We obtain

$$f_x=yz=\lambda g_x=\lambda$$ $$f_y=xz=\lambda g_y=\lambda$$ $$f_z=xy=\lambda g_z=\lambda$$

thus

$$x=y=z$$

and from $$g(x,y,z)=x+y+z=C \implies x=y=z=\frac{C}3$$

since it is a maximum point we have that

$$f(x,y,z)=xyz\le\frac{C^3}{27}\implies \frac{C}3\ge\sqrt[3]{xyz}\implies \frac{x+y+zC}3\ge\sqrt[3]{xyz}$$

To show that it is a maximum let assume for example

$$(x,y,z)=(2C/3,C/6,C/6)\implies f(x,y,z)=2\frac{C^3}{108}=\frac{C^3}{54}<\frac{C^3}{27}$$

More in general let consider z=C-(x+y) fixed thus also $x+y=k$ is fixed, from AM-GM in 2 variables we obtain

$$f(x,y,z)=xyz=xy(C-(x+y))\le k^2(C-2k)\le\frac{C^3}{27}$$

indeed it is maximum for $$k=x+y=\frac{2C}{3}$$

The same derivation can be easily generalized for more variables.