I am having trouble finding finding the KKT points of the following problem. $$ \begin{array}{ll} \min&{x^2-y^2-z^2} \\ \text{subject to} & x^4+y^4+z^4\leq1 \end{array} $$
The Lagrangian is $$ \mathcal{L}(x,\lambda)=x^2-y^2-z^2+\lambda(x^4+y^4+z^4-1) \quad \lambda\geq0 $$ KKT Conditions: $$ \begin{align} & \frac{\partial\mathcal{L}}{\partial x}=2x(1+2\lambda x^2)=0 \\ &\frac{\partial\mathcal{L}}{\partial y}=2y(-1+2\lambda y^2)=0 \\ & \frac{\partial\mathcal{L}}{\partial z}=2z(-1+2\lambda z^2)=0 \\ & x^4+y^4+z^4\leq1 \\ & \lambda(x^4+y^4+z^4-1)=0 \\ & \lambda \geq 0 \\ \end{align} $$
- If $\lambda =0$ this is easy $(x,y,z)=(0,0,0)$ is a KKT point.
- If $\lambda \neq 0$ didn't manage to solve the system. I tried substituting $x'=x^2,y'=y^2,z'=z^2$. This lead me to the following solutions: $$(x',y',z')=(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})$$ and $$(x',y',z')=(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3})$$ but because $x',y',z'$ are non-negative they don't seem to help much.
Please advise.
$\begin{align} & \frac{\partial\mathcal{L}}{\partial x}=2x(1+2\lambda x^2)=0 \\ &\frac{\partial\mathcal{L}}{\partial y}=2y(-1+2\lambda y^2)=0 \\ & \frac{\partial\mathcal{L}}{\partial z}=2z(-1+2\lambda z^2)=0 \\ \end{align}$
From first equation, $\lambda = - \dfrac{1}{2x^2}, \text{ if } x \ne 0$
Plugging into second we get $y^2 = - x^2$ so there is no solution in real.
So taking $x = 0$ from the first equation, from second equation, $\lambda = \dfrac{1}{2y^2}, \text{ if } y \ne 0$
Plugging into third equation, we get $z = \pm y$
Plugging into constraint, we get $(0, \pm \dfrac{1}{\sqrt[4]2}, \pm \dfrac{1}{\sqrt[4]2})$ at $\lambda = \dfrac{1}{\sqrt2}$
Can you take it from here?