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

Yeah so you have $x^{2}=\frac{1}{2\lambda}$ which says $x^{4}=\frac{1}{4\lambda^2}$. And you want $x^{4}+y^{4}+z^{4}=1$. So we have $$\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=1\implies \frac{3}{4\lambda^2}=1 \implies\lambda =\frac{\sqrt{3}}{2}$$ Now you have $$x^{2}=\frac{1}{2\lambda}=\frac{1}{2\cdot \frac{\sqrt{3}}{2}} =\frac{1}{\sqrt{3}}$$ Also $y^{2}=z^{2}=\frac{1}{\sqrt{3}}$ and so the maximum value is $\frac{3}{\sqrt{3}}=\sqrt{3}$.
