A Triangle Inequality

The best bound is actually given by $\frac{1 - a}{b} + \frac{1 - b}{c} + \frac{1 - c}{a} > 1.$

Since $a$ , $b$ , and $c$ are the sides of a triangle, there exist positive real numbers such that $a = y + z$ , $b = x + z$ , and $c = x + y$ . Then $2x + 2y + 2z = a + b + c = 2$ , so $x + y + z = 1$ , and $\begin{aligned} \frac{1 - a}{b} + \frac{1 - b}{c} + \frac{1 - c}{a} &= \frac{1 - y - z}{x + z} + \frac{1 - x - z}{x + y} + \frac{1 - x - y}{y + z} \\ &= \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z}. \end{aligned}$ Note that this expression is homogeneous in $x$ , $y$ , and $z$ , so we can discard the condition $x + y + z = 1$ .

We see that $\frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z} > \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z} = 1.$

Now, let $x = 1$ , $y = \epsilon$ , and $z = \epsilon^2$ , where $\epsilon$ is a positive real number. As $\epsilon \to 0^+$ , $\begin{aligned} \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z} &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon^2}{\epsilon + \epsilon^2} \\ &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon}{1 + \epsilon} \\ &\to 1, \\ \end{aligned}$ so the bound of 1 cannot be replaced by a larger number.

IGNORE THIS SOLUTION. IT IS INCORRECT.LOOK AT JON HAUSSMAN'S SOLUTION.

Because $a,b,c$ are the side lengths of the triangle, we use the well-known substitution $a=x+y$ , $b=y+z$ , and $c=z+x$ (draw a triangle and its incircle to understand why this substitution works).

Thus, we have $2x+2y+2z=2$ or $x+y+z=1$ for positive reals $x,y,z$ . Also, we have $\displaystyle\sum_{cyc}\dfrac{1-a}{b}=\sum_{cyc}\dfrac{1-x-y}{y+z}=\sum_{cyc}\dfrac{z}{y+z}$ . Thus we just need to find the minimum of $\sum_{cyc}\dfrac{y}{x+y}$

By Cauchy, we have $\left(\sum_{cyc}\dfrac{y}{x+y}\right)\left(\sum_{cyc}y(x+y)\right)\ge (x+y+z)^2=1$

Thus, $\sum_{cyc}\dfrac{y}{x+y}\ge \dfrac{1}{\sum\limits_{cyc}y(x+y)}=\dfrac{1}{x^2+y^2+z^2+xy+yz+zx}$

But $x^2+y^2+z^2+xy+yz+zx=(x+y+z)^2-(xy+yz+zx)=1-(xy+yz+zx)$ so $\dfrac{1}{x^2+y^2+z^2+xy+yz+zx}=\dfrac{1}{1-(xy+yz+zx)}$

But $xy+yz+zx\le \dfrac{(x+y+z)^2}{3}$ because when expanded and rearranged it becomes $x^2+y^2+z^2-xy-yz-zx\ge 0$ which factors into $\dfrac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge 0$ which is always true. Thus, $xy+yz+zx\le \dfrac{1}{3}$ so $\dfrac{1}{1-(xy+yz+zx)}\ge \dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}$

Thus, $\dfrac{1-a}{b}+\dfrac{1-b}{c}+\dfrac{1-c}{a}\ge \dfrac{3}{2}=\boxed{1.5}$ and we are done.

It is not reduced generalization, assume that: $a \geq b \geq c, k=\frac{1-a}{b}+\frac{1-b}{c}+\frac{1-c}{a}$ . Having: $\Leftrightarrow c^2b-c(b+a(1-a)+kab)-ab(1-b)=0$ $\Rightarrow d=(b+a(1-a)+kab)^2+4ab^2(1-b)$ $\Rightarrow 0 \leq \frac{\sqrt{(b+a(1-a)+kab)^2+4ab^2(1-b)}+(b+a(1-a)+kab)}{2b} \leq 1$ $\Rightarrow \sqrt{(b+a(1-a)+kab)^2+4ab^2(1-b)} \leq b-(a(1-a)+kab)$ $\Rightarrow b(1-ka) \gt 0 \Rightarrow k<\frac{1}{a}$ When $a \rightarrow 1\Rightarrow k \rightarrow 1\Leftrightarrow \lim_{a \rightarrow 1}{k} = 1$ $k_{max}=\boxed{1}$

As in Daniel's solution, we make the substitution $a= x+y, b= y+z, c= z+x,$ with the reverse substitution given by $x= \dfrac{b+c-a}{2}, y= \dfrac{c+a-b}{2}, z= \dfrac{a+b-c}{2}.$ The given condition transforms to $2 (x+y+z) = 2 \implies x+y+z= 1,$ and hence $\dfrac{1-a}{b} + \dfrac{1-b}{c} + \dfrac{1-c}{a} = \dfrac{x}{z+x} + \dfrac{y}{x+y} + \dfrac{z}{y+z}.$ Since the inequality is cyclic, WLOG assume $x \geq y \geq z,$ so $\dfrac{1}{y+z} \geq \dfrac{1}{z+x} \geq \dfrac{1}{x+y}.$ By rearrangement, $\dfrac{x}{z+x} + \dfrac{y}{x+y} + \dfrac{z}{y+z} \geq \dfrac{x}{x+y} + \dfrac{y}{y+z} + \dfrac{z}{z+x}.$

Note that $\left( \dfrac{x}{z+x} + \dfrac{y}{x+y} + \dfrac{z}{y+z} \right) + \left( \dfrac{z}{z+x} + \dfrac{x}{x+y} + \dfrac{y}{y+z} \right) = \dfrac{x+z}{x+z} + \dfrac{y+z}{y+z} + \dfrac{z+x}{z+x} = 3,$ so it follows that $\dfrac{x}{z+x} + \dfrac{y}{x+y} + \dfrac{z}{y+z} \geq \boxed{ \dfrac{3}{2} }.$

Equality holds for $a=b=c= \dfrac{2}{3}.$