Find the maximum value of $xy+yz+xz-2xyz$ 
If $x+y+z=1$ and $0\le x,y,z\le1$ then find the maximum value of expression $$xy+yz+xz-2xyz$$

Solution that I have
$$(1-2x)(1-2y)(1-2z)=1-2\sum x+4\sum xy-8xyz=4\sum xy-8xyz-1$$
$\implies$
$$\sum xy-2xyz=\frac{1+(1-2x)(1-2y)(1-2z)}{4}$$
Now
$$\frac{\sum(1-2x)}{3}\ge\{(1-2x)(1-2y)(1-2z)\}^{\frac13}$$
$$\implies (1-2x)(1-2y)(1-2z)\le\frac{1}{27}$$
$$\implies xy+yz+xz-2xyz\le\frac{1+\frac{1}{27}}{4}=\frac{7}{27}$$
My own solution
Consider $f(x,y,z)=xy+yz+zx-2xyz$
Since $f(x,y,z)$ is symmetric, it will achieve it's maximum value when $x=y=z=\frac13$ thereby giving $\frac{7}{27}$ as the answer.
What I want to know, is whether there is any other practical method to solve this question as considering $(1-2x)(1-2y)(1-2z)$ seems somewhat impractical and unimpressive to me. Like how will one think of such expression.
Any help is greatly appreciated.

Here is a way to use symmetry.
Using symmetry, WLOG let  $x > \max(y,z)$.  Consider replacing both $x$ and $y$ with $s = \frac{x+y}2$ in $f(x) = xy(1-2z)+(x+y)z$. The constraint is unchanged, and the value of $xy$ increases (why?). This means $f(x)$ increases, as $1-2z$ is positive in this case (why).
Hence at any maximum for this problem, we cannot have any one variable larger than the others, or the maximum can only be when $x=y=z$.  As the function $f$ is continuous and the domain is compact, we must have a maximum, viz. $f_{\max} = \frac7{27}$ when $x=y=z=\frac13$.
