Here is the original problem:
Find the extrema of $f(x,y)=xyz$ on the unit ball $xyz$ on the unit ball $x^2+y^2+z^2 \le 1$.
Here is what I got:
$f_x = yz, f_y = xz, f_z = xy$
$g_x=2x, g_y=2y, g_z =2z$
$yz = \lambda2x$
$xz = \lambda2y$
$xy = \lambda2z$
$x^2+y^2+z^2 \le 1$
=$(yz/2\lambda)^2 + (xz/2\lambda)^2 + (xy/2\lambda)^2 \le 1$
=$(yz)^2 + (xz)^2 + (xy)^2 \le 2\lambda$
=$(y^2z^2) + (x^2z^2) + (x^2y^2) \le 2\lambda$
I solved for lambda:
$xz = 2\lambda(x^2z^2)$
$1/2 = 2\lambda$
$1/4 = \lambda$
...then used lambda to solve for x in terms of yz
$yz = 2x\lambda$
$yz= 2x(1/4)$
$yz = x/2$
$2yz = x$
... and plugging it into the equation above ($xy = \lambda2z$):
$xy = 2z\lambda$
$(2yz)y = 2z(1/4)$
$2y^2z = 2z(1/4)$
= $y^2 = 1/4$
=$y = \pm 1/2$
Plugging in the rest I got ($1/2$) for x and z also. $1/2$ seems to work for the above equations but somehow this feels wrong. I tried this before and got $x= \pm1/8, y=\pm1/8, and z=\pm1/2.$ What am I doing wrong?
First, f is a monotone function of each variable in the first octant, so the maximum is on the boundary. The minimum is obtained by flipping an odd number of signs in the maximum.
With that said, you get these four relations all as equality relations. Substituting gets $xz=4\lambda^2 z/x$ which yields $x=\pm 2\lambda$ and same for $y$ and $z$, assuming none of them are zero (which is easily seen by inspection to not be the case). So each variable is $\pm 2 \lambda$, i.e. they are all the same magnitude (the actual value of $\lambda$ is not really important). Plug into the constraint to finish.
You made a mistake in your derivation in that you didn't square $2\lambda$, resulting in the wrong value of $\lambda$.