Lagrange multipliers / optimization in multivariable calculus

We want to use the multivariable calculus tools to show that, given three positive numbers a; b and c, then
(abc)^1/3 <= (a+b+c)/3
This inequality is called the arithmetic-geometric mean inequality.
(a) First, show that if x; y and z are three positive numbers satisfying xyz = 1, thenthe minimum value of the function f(x; y; z) = x + y + z is 3.
(b) Taking x = a / (abc)^(1/3), y = b/(abc)^(1/3), z = c / (abc)^1/3 deduce from (a) that the inequality holds.


I'm stuck on both parts, but I've done part of part a. This is what I've done so far:
Introducing lagrange multiplier u,
L(x, y, z, u) = x + y + z - u (xyz - 1)
let G represent the gradient vector
GL(x, y, z, u) = (1 - uyz, 1 - uxz, 1 - uxy, xyz - 1) = (0, 0, 0, 0)
so we have the followign system of equations:
1 = uyz
1 = uxz
1 = uxy
1 = xyz
I got that since none of x, y, or z can be zero:
x = y = z

From that, since xyz = 1, let c = x = y = z, c^3 = 1 so c = 1
thus x = y = z = 1 is the only critical point.
f(1, 1, 1) = 3.

This is all good, but how do I go about proving that this is a minimum and not a maximum? Would it be sufficient to show that there is another point that has a greater value, or do I need something more rigorous?


I'm also completely stuck on b.


Help would be much appreciated!

a) Keeping this as simple as possible, this point either yields a max or a min (from Lagrange).

Try a few neighboring points (x,y,z) to (1,1,1) where xyz = 1.
You will find that x + y + z > 3 in each case.

b) From part a, we know that x + y + z ≥ 3 for all (x, y, z) where xyz = 1.

Now, let x = a / (abc)^(1/3), y = b/(abc)^(1/3), z = c / (abc)^1/3; observe that xyz = 1.

So, we obtain
a / (abc)^(1/3) + b / (abc)^(1/3) + c / (abc)^(1/3) ≥ 3
==> (a + b + c)/3 ≥ 3(abc)^(1/3), as required.

I hope this helps!