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.
 A: 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 $f(3/\sqrt{2},3/\sqrt{2},0)=81/2$.
A: Here's a more routine answer. First, to find critical points in the interior, compute the gradient:
$$
\nabla f = (4x^3-4yz,4y^3-4xz,4z^3-4xy)
$$Certainly the origin is a critical point, and if any of the variables equal zero, the other two do as well. Otherwise, we have
$$
\begin{cases}
x^3-yz &=0\\
y^3-xz &=0\\
z^3-xy &=0\\
\end{cases}
$$
$$
\begin{cases}
x^4 &=xyz\\
y^4 &= xyz\\
z^4 &=xyz\\
\end{cases}
$$Since we are assuming all variables are now positive, this only admits the symmetric solution (because $x^4=y^4$ so we can infer $x=y$ and so on).
Now along the boundary. Let $f$ be the objective function and $g$ the constraint.  No harm in substituting $\lambda\mapsto 2\lambda$ to simplify the arithmetic:
$$
\begin{cases}
4x^3-4yz &=4\lambda x\\
4y^3-4xz &=4\lambda y\\
4z^3-4xy &=4\lambda z\\
\end{cases}
$$Do the same symmetry trick as before:
$$
\begin{cases}
x^4-\lambda x^2 &= xyz\\
y^4-\lambda y^2 &= xyz\\
z^4-\lambda z^2 &= xyz\\
\end{cases}
$$First, suppose $z=0$. Then either we have $(x,y)=(0,0)$, $(x,y)=(\sqrt{\lambda},0)$, or $(\sqrt{\lambda},\sqrt{\lambda})$; with $x^2+y^2+z^2=9$, these reduce to $(x,y)=(3,0)$ or $(3/\sqrt{2},3/\sqrt{2})$ (this holds up to permutation). If $xyz\neq 0$, to be quite honest, I used a computer algebra system because the equations become horrific. I couldn't find any other solutions where all variables were positive.
All that remains is to check our candidate points.


*

*$(x,y,z)=(0,0,0)$. $f(0,0,0)=0$

*$(x,y,z)=(x,x,x)$. $f(x,x,x)=3x^4-4x^3,$ $0\leq x\leq \sqrt{3}$; this has a minimum at $x=1$, corresponding to $f(1,1,1)=-1$, and a maximum of $27-12\sqrt{3}$ at $x=\sqrt{3}$

*$(x,y,z)=(3,0,0)$. $f(3,0,0)=81$.

*$(x,y,z)=(3/\sqrt{2},3/\sqrt{2},0)$. $f(3/\sqrt{2},3/\sqrt{2},0)=81/2$.


In conclusion, $f$ has a maximum at $(3,0,0)$ (cyclic) and a minimum at $(1,1,1$).
A: 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:
$$ -x = -y = -z \ \rightarrow \ x = y = z \ \ , \ \ x = -y = -z \ \rightarrow \ -x  = y = z \ \ , $$
$$ -x = y = -z \ \rightarrow \ x = -y = z \ \ \text{and} \ \ -x = -y = z \ \rightarrow \ x  = y = -z \ \ . ) $$
