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...

Let $x=b+c$, $y=a+c$ and $z=a+b$.
Thus, $a$, $b$ and $c$ are positives, $a+b+c=1$ and
$$xy+xz+yz-xyz=\sum_{cyc}(a^2+3ab)\sum_{cyc}a-\prod_{cyc}(a+b)=$$
$$=\sum_{cyc}(a^3+a^2b+a^2c+3a^2b+3a^2c+3abc)-\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)>$$
$$>\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=(a+b+c)^3=1.$$
For $a=b\rightarrow\frac{1}{2}$ and $c\rightarrow0^+$ we obtain equality,
which says that $1$ it's an infimum.
Now, for $a=b=c=\frac{1}{3}$ we obtain a value $\frac{28}{27}.$
But $$\sum_{cyc}(a^3+a^2b+a^2c+3a^2b+3a^2c+3abc)-\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\leq\frac{28}{27}$$ it's
$$\sum_{cyc}\left(a^3+3a^2b+3a^2c+\frac{7}{3}abc\right)\leq\frac{28}{27}(a+b+c)^3$$ or
$$\sum_{cyc}(a^3+3a^2b+3a^2c-7abc)\geq0,$$ which is true by Muirhead.
Thus, $\frac{28}{27}$ it's a maximal value.
Since our expression is continuous, we got the answer: $$\left(1,\frac{28}{27}\right].$$
