Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ . 
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .

What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$ 
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.
But I think it is a wrong method to proceed...

$xy+yz+xz-xyz=(1-x)(1-y)(1-z)+x+y+z-1=(1-x)(1-y)(1-z)+1$.
And $x+y+z=2$ gives $(1-x)+(1-y)+(1-z)=1$.
$1-x,1-y,$ and $1-z$ are all strictly positive (otherwise we couldn't make a triangle out of $x,y,$ and $z$). So let $a=1-x,b=1-y,$ and $c=1-z$. We have
$$a,b,c\in(0,1)$$
$$a+b+c=1$$
From $AM\ge GM$, their product satisfies $abc\in(0,\frac{1}{27}]$. You can take it from there.
