You want a lower bound of the probability of $[X\gt x]$ hence an upper bound of the probability of the event $A=[X\leqslant x]$. As explained by others there is little hope to achieve such a bound depending on $\mathrm E(X)$ only, which would be valid for every nonnegative random variable $X$.
However, for every decreasing bounded function $u$, $A=[u(X)\geqslant u(x)]$ hence Markov's inequality yields $$ \mathrm P(A)\leqslant u(x)^{-1}\mathrm E(u(X)). $$ Two frequently used cases are $$u(x)=\mathrm e^{-tx}$$ and $$u(x)=\frac1{1+tx}$$ for some positive $t$, related to Laplace and Stieltjes transforms, respectively. In both cases, one can choose the value of the parameter $t$ which yields an optimal, or nearly optimal, upper bound.
This yields $$ \mathrm P(X\gt x)\geqslant 1-u(x)^{-1}\mathrm E(u(X)). $$ A simple consequence is the fact that, for every positive $s$ (and for $s=0$ as well, provided $1/X$ is integrable), $$ \mathrm P(X\gt x)\geqslant \mathrm E\left(\frac{X-x}{s+X}\right). $$