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?
 A: 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.
A: Proof:
Let
$$x = \frac{a + b - c}{2} > 0, \quad y = \frac{b + c - a}{2} > 0, \quad z = \frac{c + a - b}{2} > 0.$$
Correspondingly, $a = z + x, b = x + y, c = y + z$ (the so-called Ravi's substitution).
We have $a + b + c = 2(x + y + z)$. Thus, $x + y + z = 1$.
Also, we have
\begin{align*}
 &ab + bc + ca - abc \\
 =\,& (z + x)(x + y) + (x + y)(y + z) + (y + z)(z + x) - (z + x)(x + y)(y + z)\\
 =\,& (z + x + y + z)(x + y) + (y + z)(z + x)(1 - x - y)\\
 =\,& (1 + z)(1 - z) + [xy + z(x + y + z)]z \\
 =\,& 1 - z^2 + (xy + z)z \\
 =\,& 1 + xyz.
\end{align*}
First, $ab + bc + ca - abc = 1 + xyz > 1$.
Second, using AM-GM, we have $xyz \le (\frac{x + y + z}{3})^3 = \frac{1}{27}$ with equality if and only if $x = y = z = 1/3$.
Thus, we have $ab + bc + ca - abc \le \frac{28}{27}$ with equality if and only if $a = b = c = 2/3$.
We are done.

Remark: Actually, we can prove the following identity:
\begin{align*}
 &ab + bc + ca - abc\\
 \equiv\,& 1 + \frac{(a + b - c)(b + c - a)(c + a - b)}{8}\\
 & + \frac{1}{8}(a + b + c - 2)(a^2 + b^2 + c^2 - 2ab - 2bc - 2ca + 2a + 2b + 2c + 4).
\end{align*}
