Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$
$\nabla f_x$ $=$ $2xy^{2}z^{2}$, 
$\nabla f_y$ $=$ $2yx^{2}z^{2}$, 
$\nabla f_z$ $=$ $2zx^{2}y^{2}$
$\nabla g_x$ $=$ $2x$, 
$\nabla g_y$ $=$ $2y$, 
$\nabla g_z$ $=$ $2z$
setting the sets equal to each other and multiplying each to get the equality where $\nabla f_{x..z} = 2x^2y^2z^2$ gives me(namely multiplying $x$ on $\nabla f_x = \lambda g_x$...and so on) :
$\lambda2x^2 = \lambda2y^2=\lambda2z^2 \to x=y=z$ after division and taking the roots. So:
$x=y=z=a \to 3a^2=1 \to a= \pm \frac{1}{\sqrt{3}}$
Thus a maximum at $f\Big(\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}\Big) = \frac{1}{27}$.
However looking at wolrfram alpha, it also gives a minimum at $f\Big(-\frac{1}{2},-\frac{\sqrt{3}}{2},0\Big) = 0$. I would like to know how there are more solutions from $\lambda 2x^2=\lambda 2y^2=\lambda 2z^2$ than just x=y=z
Using the corrections from posters lets say I let $x=0$ now I have $y^2+z^2=1$ Then if I wanted one scenario just to show there is a minimum I could say $y^2=z^2$ and then say $a^2+a^2=1 \to 2a^2=1 \to a^2=\frac{1}{2} \to a=\pm\frac{1}{\sqrt{2}}$ therefore have a minimum at $f\Big(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)$ Is this correct. I feel the only correct solution thought is to use set notation and proof to show there are infinitely many cases. But does my above asnwer for one of those cases hold?
 A: You have three equations from the first order conditions:
$$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$
Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This gives us infinitely many solutions with $\lambda=x=0$, since we can choose any $y$ and $z$ such that $y^2+z^2=1$. (Geometrically, these solutions are just intersections of the constraint set - a sphere of radius $1$ centred at the origin - and the $x=0$ plane. The intersection is a circle of radius $1$ centred at the origin in the $x=0$ plane.) Mathematically, the set of such minimizers is:
$$\{(0,y,z)\in\mathbb{R}^3\ \ |\ \ y^2+z^2=1\}$$
Similar arguments hold for the cases $y=0$ and $z=0$ (everything is symmetric). These critical points are all minimizers of $f$ on the constraint set. The minimum is $0$. (Notice that $\lambda=0$ here because even if we did not have the equality constraint, these points would minimize $f$.)
You lost all these solutions because you inadvertently assumed all the variables were nonzero when you divided the equations by one another.

Now suppose that $x$, $y$ and $z$ are all nonzero. Then $\lambda\neq0$ (this is implied by any of the three equations). Now since all the variables are nonzero, we can divide the equations by $2x$, $2y$ and $2z$ respectively to get
$$y^2z^2=\lambda  \qquad x^2z^2=\lambda \qquad x^2y^2=\lambda.  $$
Then dividing each equation by each other equation we find that we require $$x^2=y^2=z^2.$$
Plugging into the constraint then implies that $$x^2=y^2=z^2=\frac{1}{3}$$
and $\lambda =1/9$ (from any of the three equations). This gives us $8$ solutions, which are maximizers of $f$ on the constraint set. The maximum is $1/27$.
A: You didn't consider the case $x = y = \lambda = 0, z = 1$  which satisfies the nabla equations, and gives the minimum value for $f$ of $0$.
