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...

Given any three positive numbers $x,y,z$, it can form the sides of a non-degenerate triangle if and only if we can find three positive numbers $u, v, w$ such that
$$x = v +w,\quad y = u + w\quad\text{ and }\quad z = u +v$$
(This is known as the Ravi's substituion).
In terms of them, the condition $x + y + z = 2$ is equivalent to $u + v + w = 1$.
Under this condition, we have
$$\begin{cases}
u &= 1 - (v+w) = 1 - x\\
v &= 1 - (u+w) = 1 - y\\
w &= 1 - (u+v) = 1 - z
\end{cases}$$
Notice
$$(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz$$
The expression at hand equals to
$$\begin{align}xy+yz+zx - xyz 
&= (1-x)(1-y)(1-z) - 1 + (x+y+z)\\
&= uvw - 1 + 2 = 1+uvw
\end{align}$$
By AM $\ge$ GM, we know $uvw \le \left(\frac{u+v+w}{3}\right)^3 = \frac1{27}$. Since $u,v,w > 0$, we have
$$1 < xy+yz+zx - xyz \le \frac{28}{27}$$
for any valid choice of $x,y,z$.
In fact, if we set $(u,v,w)$ to $(t,t,1-2t)$ and let $t$ increases from $0$ to $\frac13$, the expression $1+uvw$ increases from $1$ continuously to $\frac{28}{27}$. From this, we can deduce the range of the expression $xy + yz + xz - xyz$ is $(1,\frac{28}{27}]$.
