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.
 A: 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 $(A, B, C, D)$.
A: Let $\mathcal{T}$ be the regular tetrahedron of side $\sqrt{2}$ with vertices at
$\begin{cases}
A &= (+1,+1,+1),\\
B &= (+1,-1,-1),\\
C &= (-1,+1,-1),\\
D &= (-1,-1,+1).
\end{cases}$
For every point $X \in \mathcal{T}$, its baricentric coordinate is an unique $4$-tuple $(a,b,c,d)$ such that
$$\begin{cases}
a, b, c, d \ge 0, a + b + c + d = 1\\
X = aA + bB + cC +dD
\end{cases}
$$
Conversely, any $4$-tuple $(a,b,c,d)$ satisfies the first condition
define a point $X$ on $\mathcal{T}$ by the second formula. 
Let $X = (x,y,z)$, it is easy to work out
$$\begin{cases}
a &= \frac14 (+x + y + z + 1)\\
b &= \frac14 (+x - y - z + 1)\\
c &= \frac14 (-x + y - z + 1)\\
d &= \frac14 (-x - y + z + 1)
\end{cases}$$ 
In terms of $(x,y,z)$, we have
$$abc+bcd + cda + dab - abcd = \frac{15-M(x,y,z)}{256}$$
where 
$$M(x,y,z) = (14-(x^2+y^2+z^2))(x^2+y^2+z^2)+2(x^4+y^4+z^4)-24xyz$$
Let $r^2 = x^2 + y^2 + z^2$. Using $AM \ge GM$, we have
$$|xyz| \le \left(\frac{x^2+y^2+z^2}{3}\right)^{3/2} = \frac{r^3}{3\sqrt{3}}
\implies
M(x,y,z) \ge (14-r^2)r^2 - \frac{8r^3}{\sqrt{3}}
$$
As $X = (x,y,z)$ varies over $\mathcal{T}$, $r$ take values from the interval
$[0,\sqrt{3}]$. It is easy to check the polynomial on RHS is non-negative over
such a interval. This means $M(x,y,z) \ge 0$ for $X \in \mathcal{T}$.
As a result, we have
$$abc + bcd + cda + dab - abcd \le \frac{15}{256}$$
Since $M(0,0,0) = 0$, the equality is achieved at $a = b = c = d = \frac14$. This implies the maximum value we seek is $\frac{15}{256}$.
A: The following solution is similar to my answers
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
and An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $x = a + b, \, y = c + d$.
We have $x + y = 1$.
We have
\begin{align*}
 M &= ab y + cd x - ab cd\\
 &= xy - (x - ab)(y - cd)\\
 &\le xy - (x - x^2/4)(y - y^2/4) \tag{1}\\
 &= -\frac{1}{16}x^2y^2 + \frac{1}{4}xy(x + y)\\
 &= -\frac{1}{16}x^2y^2 + \frac{1}{4}xy\\
 &= -\frac{1}{16}(2 - xy)^2 + \frac14 \\
 &\le -\frac{1}{16}(2 - 1/4)^2 + \frac14 \tag{2}\\
 &= \frac{15}{256}
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4 \le x$ and $cd \le (c + d)^2/4 = y^2/4 \le y$ in (1), and
$xy \le (x + y)^2/4 = 1/4$ in (2).
Also, when $a = b = c = d = 1/4$,
we have $a + b + c + d = 1$ and $M = 15/256$.
Thus, the maximum of $M$ is $15/256$.
