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.

Another way for proving that $f(x,y,z)=xy+yz+zx-2xyz\leq \frac{7}{27}$ if $x,y,z\in(0,1)$ and $x+y+z=1.$ We denote $x=X+1/3,y=Y+1/3, z=Z+1/3$ with $X,Y,Z\in [-1/3.2/3]$ and $X+Y+Z=0.$ We denote
$$A=-2(XY+YZ+ZX)=X^2+Y^2+Z^2$$ leading to
$$f(x,y,z)=\frac{7}{27}-\frac{1}{6}(A+12XYZ).$$ If 2 of the three numbers $X, Y,Z$ are negative we have $A+12XYZ\geq 0.$ If not, without loss of generality we assume $Y$ and $Z$ non negative. As a consequence $0\leq Y+Z=-X\stackrel{(*)}
{\leq} 1/3$ and we get
$$A+12XYZ=(Y+Z)^2+Y^2+Z^2-12 (Y+Z)YZ\stackrel{(*)}
{\geq }2(Y^2+Z^2+YZ) -4YZ=2((Y-Z)^2+YZ)\geq 0.$$
