I have this multivariate polynomial function
$$\begin{align*} f(L,x,y,z) &= L^2 + (1 - L x)^2 + (1 - L y)^2 + (1 - L z)^2 \\ &+ (2 - L x y)^2 + (2 - L x z)^2 + (2 - L y z)^2 + (3 - L x y z)^2 \end{align*}$$
To compute its minimum an idea is to computes its critical points and then verify which of them minimizes $f$. This is a differentiable function, so all maxima and minima of $f$ are critical points, right?
Solving the equation $\nabla f(L,x,y,z) = (0,0,0,0)$ we can find its critical points. Supposing $L \neq 0$, we have the following polynomial system. $$ \begin{cases} \begin{align*} X &+ Y + Z + 2 X Y + 2 X Z + 2 Y Z + 3 X Y Z \\ &-L(1 + X^2 + Y^2 + Z^2 + X^2 Y^2 + X^2 Z^2 + Y^2 Z^2 + X^2 Y^2 Z^2) = 0 \end{align*}\\ -1 + Y - Y^2 + Z + 2 Y Z - Z^2 - Y^2 Z^2 = 0\\ -1 + X - X^2 + Z + 2 X Z - Z^2 - X^2 Z^2 = 0\\ -1 + X - X^2 + Y + 2 X Y - Y^2 - X^2 Y^2 = 0\\ \end{cases}$$
I tried to solve this system using Mathematica and PHCPack. Both programs solved the system without a problem, but both missed the actually minimum. I had to use a minimizer routine in Mathematica to get the minimum $(L,x,y,z) = \left( \frac{1+\sqrt{17}}{9}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4} \right)$. This point minimizes $f$ but is not a solution of the system above.
My question is, why this minimum is not a critical point? How is that possible?
