Maximum Value satisfying a variation of the Triangle Inequality The Triangle Inequality states that:
$$||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$$
Now, suppose that we have a triangle with side lengths of $a,b,c$ such that $a+b+c=2$. Avoiding any assumptions of what kind of triangle we have, what the angle measures are, etc., 
Now suppose we have the following inequality:
$$k\le\dfrac{1-a}{b}+\dfrac{1-b}{c}+\dfrac{1-c}{a}$$
My question is that what can $k$ be. Meaning what is the maximum possible $k$ which satisfies the inequality I gave. I just don't want an answer, but a proof as to what the maximum value of $k$ will be.

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

Proof:
Since $a$, $b$, and $c$ are the sides of a triangle, there exists positive real numbers such that $a = y + z$, $b = x + z$, and $c = x +y$. Therefore, $2x + 2y + 2z = a + b + c = 2\implies x + y + z = 1$, and
  $$ \begin{align*} \frac{1 - a}{b} + \frac{1 - b}{c} + \frac{1 - c}{a} &= \frac{1 - y - z}{x + z} + \frac{1 - x - z}{x + y} + \frac{1 - x - y}{y + z} \\ &= \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z}. \end{align*} $$
  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 some positive real number. As $\epsilon \to 0^+$,
  $$ \begin{align*} \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z} &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon^2}{\epsilon + \epsilon^2} \\ &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon}{1 + \epsilon} \\ &\to 1, \\ \end{align*} $$
  so the bound of $1$ cannot be replaced by a larger number.

