We complete the square. Factoring $ 4 $ out of the quadratic and linear terms gives $ 4x^2 + 2x = 4\left(x^2 + \frac12x\right) $. Since $ \left(x+\frac14\right)^2 = x^2 + \frac12x + \frac1{16} $, we can write $$4\left(x+\frac14\right)^2 = 4x^2 + 2x + \frac14.$$This quadratic agrees with the given $ 4x^2+2x-1 $ in all but the constant term. We can write \begin{align*} 4x^2 + 2x - 1 &= \left(4x^2 + 2x + \frac14\right) - \frac 54 \\ &= 4\left(x+\frac 14\right)^2 - \frac 54.\end{align*}Therefore, $ a=4 $, $ b=\frac14 $, $ c=-\frac54 $, and $ a+b+c = 4+\frac14-\frac 54 = \boxed{3} $.