<s>[INST] 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.
 [/INST] 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 $\frac{15}{256}$.
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