Prove $1If $a$, $b$, $c$ are three positive real numbers such that:
$a+b>c$
$b+c>a$
$c+a>b$
$a+b+c=2$
$a$, $b$ and $c$ might be or might not be equal in value
Show that:
$$1<ab+bc+ca-abc<\frac{28}{27}$$
From the first half of the question, I realised that $a$, $b$ and $c$ are the sides of a triangle whose perimeter is $2$. However, I'm unable to solve this problem, even after using some standard inequalities.
Can someone help me with this problem?

I'll write up my comment as an answer. Consider the polynomial
$$
\begin{split}
f(x)&=(x-a)(x-b)(x-c)\\
&=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc\\
&=x^3-2x^2+(ab+bc+ca)x-abc.
\end{split}
$$
Then $f(1)=1-2+ab+bc+ca-abc=ab+bc+ca-abc-1$, so we need to show that
$0<f(1)\le 1/27$.
Note that $1=(a+b+c)/2=p$, the semiperimeter of the triangle with sides $a,b,c$. This triangle is not degenerate since all the inequalities in the problem statement are strict. Therefore, its area is positive. But the area of this triangle is
$$
A=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{1\cdot f(1)}=\sqrt{f(1)},
$$
so $f(1)=A^2>0$. On the other hand, by the AM-GM inequality,
$$
\begin{split}
\sqrt[3]{f(1)}&=\sqrt[3]{(p-a)(p-b)(p-c)}\\
&\le \frac{(p-a)+(p-b)+(p-c)}{3}\\
&=\frac{3p-(a+b+c)}{3}=\frac{3-2}{3}=\frac{1}{3},
\end{split}
$$
i.e. $f(1)\le (1/3)^3=1/27$.
In fact, we can use AM-GM inequality on any triangle to show that its area $A\le\dfrac{p^2}{3\sqrt{3}}$, where $p$ is its semiperimeter.
