<s>[INST] 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?
 [/INST] 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 $0$.
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