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.