If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$

The continuous function $$ f(a, b, c, d) = abc+abd+acd+bcd-abcd $$ has a maximum on the compact set $$ \{ (a, b, c, d) \in \Bbb R^4\mid a, b, c, d \ge 0, a+b+c+d = 1 \} $$ which is obtained at some point $(A, B, C, D)$. If we can show that $A=B=C=D$ then it follows that the maximum is $$ M = f(\frac 14,\frac 14,\frac 14,\frac 14) = \frac{15}{256} \, . $$

First observe that at most one of $(A, B, C, D)$ can be zero, since otherwise $f(A, B, C, D) = 0$, and $0$ is not the maximum.

Now assume that $A=B=C=D$ does not hold. Without loss of generality, assume that $A \ne B$. From $$ f(a, b, c, d) = ab(c+d-cd) + (a+b)cd $$ it follows that $$ f(\frac{A+B}2, \frac{A+B}2,C, D) - f(A, B, C, D) = \left( \frac{(A+B)^2}4 - AB \right) (C + D - CD) \\ = \frac{(A-B)^2}4 (1 - (1-C)(1-D)) $$ and that expression is $> 0$ because $A \ne B$, and $C, D$ are not both zero. So we have $$ f(\frac{A+B}2, \frac{A+B}2,C, D) > f(A, B, C, D) \, , $$ contradicting the assumption that $f$ obtains its maximum at $(A, B, C, D)$.