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

I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$. 
Thanks.

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 $(A, B, C, D)$.
