If $x+y+z+a+b+c=1$, $xyz+abc=1/36$, what is the max value of $abz+bcx+cay$? Question : If $x+y+z+a+b+c=1,xyz+abc=1/36,$ What is the max value of $abz+bcx+cay$?  ($x, y, z, a, b, c ∈ R$ (non-negative))
I tried $abz+bcx+cay≥3abc$ (because of AM-GM) to make it similar to $xyz+abc=1/36$, but I'm stuck with expanding.
 A: For the first equation, we can test the waters with a naive solution
$x+y+z+a+b+c=1\qquad=6\left(\dfrac16\right)
=\dfrac16+\dfrac16+\dfrac16+\dfrac16+\dfrac16+\dfrac16 \tag{1}$
and it appears that at least $\space 2\space $ of $\space a,b,c,x,y,z$ must be other than $\space \frac16\space $ for the other equation to work. Again, we suggest a non-working naive solution.
$$\text{So we let}\quad xyz+abc=\dfrac{1}{36}
=\frac{1}{2\cdot6\cdot6}+\frac{1}{2\cdot6\cdot6}???\\
\implies 36(xyz+abc)=1\\
\implies x+y+z+a+b+c=36(xyz+abc)\tag{2}$$
$$\implies
\big( a + b + c -36 a b c\big) +\big(x + y + z - 36 x y z\big)=0\\
\tag{3}$$
Assuming that both addends in $(3)$ equal zero.
$$( a + b + c - 36 a b c)  = 0\\
\implies c = \frac{a + b}{36 a b - 1}\\
\implies 36 a b - 1>0 \implies b>\frac{1}{36a}\tag{4}
$$
I found $\space \big(a=1/12\quad b=1/3\quad c=1/2  \quad x=y=z=1/36\big)\space $ for equation $(1)$ with a,b interchangeable but a,b violate the condition of $\space b>\dfrac{1}{36a}$ and the x,y,z values of $1/36$ do not work with the the right part of equation $(3)$.  This means that the two parts of equation $(3)$ are not separately zero an combinations must make them equal but with opposite signs.
Forgive me but it is past my bedtime and I cannot finish this tonight.
I can suggest is to trying the $2,3,4,6,9,12,18,36$- factors of $36$ as denominators to see which add up to $\space 1\space $ and then seeing which satisfy the second equation. From there,  you can select the combinations that yield the highest value for the final equation in your question.
Perhaps an even faster approach would be to use equation $(4)$ equal to a non-zero factor of $36$ and  test when the swapping of $a,b$ yields an opposite sign. That is the approach I would take if I did not have to get to bed. Good luck.
