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
 A: $x^2+y^2+z^2 \ge x^4+y^4+z^4 = 1$, and equality occurs when $x = 0,y = 0, z = \pm 1$ or permutations of them. Also by Cauchy-Schwarz inequality: $x^2+y^2+z^2 \le \sqrt{3(x^4+y^4+z^4)} = \sqrt{3}$ with equality occurs when $x = y = z = \pm \dfrac{1}{\sqrt[4]{3}}$ . Thus we can conclude that the min $= 1$ and the max $= \sqrt{3}$ .
A: 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.
A: 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 $\frac{3}{\sqrt{3}}=\sqrt{3}$.
