<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] 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 $(1,1,1$).
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