AM-GM inequality only!

The function $\frac{1}{x(x+1)}$ approaches infinity near $x = 0$ and decreases diminishingly, so naturally a, b, and c must each be as far away from 0 as possible, therefore $a = b = c = \frac{1}{3}$ . I also want to mention that the problem should specifiy that a, b, and c must be positive, because with very large negative numbers and and positive numbers that add up to 1, the value approaches zero

Here's my solution:

Let $f(x)=\frac{1}{x(x+1)}$ . Since $f"(x)=\frac{1}{[a(a+1)]^4}$ , $f(x)$ is a convex function. And hence, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ so, we have $\sum_{cyc} \frac{1}{a(a+1)} \geq \frac{27}{4}=\boxed{6.75}$