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.

Integrand's concern about the case for $ \ xyz \ \ne \ 0 \ $ turns out to be unnecessary.  If we solve the original Lagrange equations for $ \ \lambda \ $ , we obtain
$$ \left\{ \begin{array}{c}
2·(x^3-yz)  \ = \ \lambda x \\ 2·(y^3-4xz)  \ = \ \lambda y \\
2·(z^3- xy) \ = \ \lambda z
\end{array} \right. \ \ \Rightarrow \ \ \lambda \  =  \ \frac{2·(x^3-yz)}{x} \ =  \ \frac{2·(y^3-xz)}{y} \ =  \ \frac{2·(z^3-xy)}{z} \ \ . $$
Cross-multiplying the first ratio-pair produces
$$ x^3y \ - \  y^2z \ \ = \ \ xy^3  -  x^2z \ \ \Rightarrow \ \ x^2z \ -  \ y^2z \ \ = \ \ xy^3 - x^3y $$ $$ \Rightarrow \ \ z·(x^2-y^2) \ \ = \ \ xy·(y^2 - x^2) \ \ \Rightarrow \ \  (z \ + \ xy) \ · \ (x^2-y^2) \ \ = \ \ 0  \ \ ;   $$
similarly, we have from the other ratio-pairs,
$$ (y \ + \ xz) \ · \ (x^2-z^2) \ \ = \ \ 0  \ \ \ , \ \ \ (x \ + \ yz) \ · \ (y^2-z^2) \ \ = \ \ 0  \ \ .   $$
[That each of these equations looks like a "cyclic-permutation" re-labeling of the others is due to the same symmetry being found in the expression for $ \ f(x,y,z) \ $ and the symmetry of the spherical constraint.]
The "difference-of-squares" factors would tell us such things as $ \ y = \pm x \ , $ but since we are also restricted to the first octant, we have just $ \ x = y = z \ $ as one case.  This gives the $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ results.
The remaining case comes from the factors such as $ \ z + xy \ = \ 0 \ \Rightarrow \  z \ = \ -xy \ . $  Again because we are in the first octant, including those portions of the coordinate planes, this requires that $ \ z = 0 \ $ and either $ \ x = 0 \ \ \text{or} \ \ y=0 \ . $  Together with the related results from the other two equations, we obtain the three critical points which are the permutations of $ \ (3 \ , \ 0 \ , \ 0 ) \ $  and the three permutations of $ \ (3\sqrt{2} \ , \ 3\sqrt{2} \ , \ 0 ) \ . $  So it does not appear that any critical points have been overlooked and the conclusion in Integrand's 19 May post is confirmed.
(Incidentally, the origin is a rather interesting saddle point.  The function $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ has an inflection point at $ \ x = 0 \ $ , so this change in concavity occurs along each of the lines passing between opposing octants:
$$ -x = -y = -z \ \rightarrow \ x = y = z \ \ , \ \ x = -y = -z \ \rightarrow \ -x  = y = z \ \ , $$
$$ -x = y = -z \ \rightarrow \ x = -y = z \ \ \text{and} \ \ -x = -y = z \ \rightarrow \ x  = y = -z \ \ . ) $$
