Let $X$ have finite variance and $b>0$. I want to show $$P(X \ge E(X) + b) \le \frac{V(X)}{b^2+V(X)}$$The exercise gives a hint: first show that $$(b+t)^2P(X\ge E(X)+b)\le V(X)+t^2$$ for $t\ge 0$.
I've looked for inequalities on the probability of $X\ge E(X)$ and the like, but I have found none that seem to fit this problem. Chebyshev's inequality for instance relates to the absolute value.
Is there a tool I don't know to break this probability down? Or does anyone see a cleverer way?