Prove $1<ab+bc+ca-abc<\frac{28}{27}$

Question

If $a$, $b$, $c$ 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?

Answer 1

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.

Answer 2

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.

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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*}