Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+z^4$ and the last equation is equivalent to the system of equations $$\begin{cases}
2x=4\lambda x^3 \\
2y=4\lambda y^3 \\
2z=4\lambda z^3
\end{cases}$$
After dividing into $2$ and multiplying to $x,y$ and $z$, respectively we get: $$\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}$$
Considerong the first equation we get two cases: $x=0$ or $1-2\lambda x^2=0$
After that I am stuck. How to rule out or consider each case?
Can anyone demonstrate it clearly?
Would be very thankful for help

You have
\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}
Case 1: Assume $xyz\ne 0.$ Then $x^2=y^2=z^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{3}{4\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Case 2: Assume $z=0,xy\ne 0.$ Then $x^2=y^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{1}{2\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Case 3: Assume $y=z=0,x\ne 0.$ Then $x^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{1}{4\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Note that $x=y=z=0$ is not possible and that, because of symmetry, case $2$ cover all possibilities with one variable zero and case $3$ cover all possibilities with two variables zero.
