How find this maximum $f(a,b,c,x,y,z)=(a-b)z+(b-c)x+（c-a)y$

Question

let $a,b,c,x,y,z\ge 0$,and $a+b+c=x+y+z=1$,find this maximum $$f(a,b,c,x,y,z)=(a-b)z+(b-c)x+(c-a)y$$ Thank you everyone

Answer 1

This is a geometric solution: Consider $(a,b,c),(x,y,z)$ are two points on the plane $x+y+z=1$ with normal vector $(1,1,1)$. (hope the notation is not confusing here.) Notice that $$(b-c,c-a,a-b)=(a,b,c)\times(1,1,1)$$is a vector that is parallel to the plane $x+y+z=1$. Since $$f(a,b,c,x,y,z)=(b-c,c-a,a-b)(x,y,z),$$and $(x,y,z)$ is a vector starting from the origin pointing to a point on the plane, so $f$ can reach the max if and only if $(b-c,c-a,a-b)$ and $(x,y,z)$ have the minimum angle between.

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The above is the geometric interpretation, here is the more straightforward answer:

$$f(a,b,c,x,y,z)=(x,y,z)\cdot((a,b,c)\times(1,1,1))$$reaching the max at their volumn of the solid generated by them.

Answer 2

The maximum is $1$ and achieved for example if $a=z=1, b=c=x=y=0$. Note that $f(a,b,c,x,y,z)$ is a convex combination of $a-b, b-c,c-a$, hence cannot exceed $\max\{a-b, b-c,c-a\}$.