Proof concavity/convexity on $f(x,y,z)=1+x+xy+xyz$

Question

Say $f(x,y,z)=1+x+xy+xyz$

How do I prove that $f$ is concave on $x,y,z\in[0,1]$?

Answer 1

It's neither convex nor concave.

Computing the Hessian matrix gives us $$Hf = \begin{bmatrix} 0 & 1+z & y \\ 1+z & 0 & x \\ y & x & 0\end{bmatrix}.$$ This is indefinite for almost all values of $x$, $y$, and $z$. Since the trace is $0$, the eigenvalues must sum to $0$, so either there are both positive and definite eigenvalues (making the matrix indefinite) or else all three eigenvalues are $0$.

Since the matrix is diagonalizable (it's symmetric) this can only happen when it's the zero matrix: when $(x,y,z) = (0,0,-1)$. It's sort of meaningless to say that a function is convex or concave at a single point, though, and $(0,0,-1)$ is outside the set you're interested in anyway.

Answer 2

Let $\{a,b,c\}\subset[0,1]$.

Since $$\frac{\partial f}{\partial x}=1+y+yz,$$ $$\frac{\partial f}{\partial y}=x+xz$$ and $$\frac{\partial f}{\partial z}=xy,$$ we need to prove for all $\{x,y,z\}\subset[0,1]$ that $$(1+b+bc)(x-a)+(a+ac)(y-b)+ab(z-c)\geq f(x,y,z)-f(a,b,c)$$ or $$bx+bcx+ay+acy+abz\geq ab+2abc+xy+xyz,$$ which is a linear inequality of $x$, of $y$ and of $z$.

Thus, it remains to prove the last inequality for $\{x,y,z\}\subset\{0,1\}$,

which is wrong for $x=y=z=0$.