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.

Suppose that $a\geq b\geq c$, so $1-a\leq 1-b\leq 1-c$ and $\frac{1}{a}\leq \frac{1}{b}\leq \frac{1}{c}$. By Tchebyshev inequality you have
$$\frac{1-a}{b}+\frac{1-b}{c}+\frac{1-c}{a}\geq 3\cdot \frac{1}{3}(1-a+1-b+1-c)\cdot \frac{1}{3}(1/a+1/b+1/c)=\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}).$$
By Cauchy-Schwarz you have $(a+b+c)(1/a+1/b+1/c)\geq 9$ so $(1/a+1/b+1/c)\geq 9/2$.
Then $k=\frac{3}{2}$ and its obtain for $a=b=c=2/3$.
