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...
 A: 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 $(1,\frac{28}{27}]$.
A: 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: $$\left(1,\frac{28}{27}\right].$$
A: $xy+yz+xz-xyz=(1-x)(1-y)(1-z)+x+y+z-1=(1-x)(1-y)(1-z)+1$.
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.
A: Let $F$ denote the function $xy + yz + xz - xyz$. Since $x + y + z = 2$, the area of the triangle by Heron's formula is
$$A = \sqrt{(1-x)(1-y)(1-z)} = \sqrt{F - 1}.$$
Hence the minimum value of $F$ is $1$, which is attained iff we allow triangles of zero area.
The maximum $F$ occurs for maximum $A$. If we're allowed to assume that the triangle with maximum area for a given perimeter is equilateral, the maximum $F$ occurs when $x = y= z= 2/3$, which gives $F = 28/27$ as others have found. If not, we can prove it as follows. Fix $x$ and write
$$F = yz(1 - x) + x(2 - x).$$
By the triangle inequality $x \le 1$, so $F$ is a maximum when $yz$ is a maximum. Since $y + z$ is fixed, the maximum $F$ for this $x$ occurs when $y = z$. Similarly for a fixed $y$ and a fixed $z$. Hence the maximum $F$ occurs for an equilateral triangle.
