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
Best Answer
$x^2+y^2+z^2 \ge x^4+y^4+z^4 = 1$, 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 $= \sqrt{3}$ .