# Markov inequality: More precise bound?

Given a random variable X, we know that $$P[X\geq A] = 1$$. By Markov inequality, we obtain that $$E[X]\geq A$$. Or, in other words, $$E[X] = A + \lambda,\,\,\lambda\geq 0$$. Is there any way I can more precisely characterize the $$\lambda$$? E.g., if I know the variance of $$X$$? Or applying some other bounds, less conservative than Markov’s.