<s>[INST] Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theorem, by parametrizing the sphere and by algebraic manipulating the function but I haven't been able to come up with the solution.
Could anybody help me?
Thank you in advance.
 [/INST] Introduce the dummy variable $w$. Then we want to minimize
$$
F(x,y,z,w)=x^4+y^4+z^4+w^4-1-4xyzw
$$subject to $x^2+y^2+z^2\leq 9, w=1$. By AM-GM
$$
(x^4+y^4+z^4+w^4)\geq4 \sqrt[4]{x^4y^4z^4w^4}=4xyzw,
$$with equality iff $x=y=z=w$. Then since we must have $w=1$, the minimum occurs at $(x,y,z)=(1,1,1)$, giving $f(1,1,1)=-1$.
Now for the maximum. Note that
$$
f(x,y,z) = \langle x^2,y^2,z^2\rangle \cdot \langle x^2,y^2,z^2\rangle-4xyz
$$By Cauchy-Schwarz,
$$
|f(x,y,z)|\leq (9)^2-4xyz;
$$equality occurs if any of the variables are $0$. At $(3,0,0)$ we have $f(3,0,0)=81$ and at $(3/\sqrt{2},3/\sqrt{2},0)$, we have $f(3/\sqrt{2},3/\sqrt{2},0)=81/2$.
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