Probability of a random variable greater than its expected value We have a lot of probabilities lower bounding as  (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality)
$$
P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1
$$
However, It would be great to know if there is any inequality bounding exactly
$$
P(X > E[X]) \geq c
$$
i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)
 A: Let $Y:=X-EX$. We need to obtain a lower bound on $P(Y>0)$. 
Suppose that $-a\le Y\le b$ for some real $a>0$ and $b>0$, and that $EY^2\ge s^2$ for some real $s$. Then 
$$1_{Y>0}\ge\frac{aY+Y^2}{ab+b^2}.$$
Taking expectations of both sides of this inequality, we get 
$$P(Y>0)\ge\frac{s^2}{ab+b^2}. \tag{1}$$
In terms of $X$, (1) can be rewritten as 
$$P(X>EX)\ge\frac{Var\,X}{ab+b^2},$$
provided that $-a\le X-EX\le b$.

The condition $-a\le Y\le b$ implies that 
$$Y^2\le\frac{Y+a}{a+b}\,b^2+\frac{b-Y}{a+b}\,a^2.$$
Taking expectations of both sides of this inequality, we get 
$$s^2\le EY^2\le\frac{a}{a+b}\,b^2+\frac{b}{a+b}\,a^2=ab.$$
So, letting now 
$$p:=\frac{s^2}{(a+b)a}\quad\text{and}\quad r:=\frac{s^2}{(a+b)b}, $$
we see that 
$$p+r=\frac{s^2}{ab}\le1.$$
Letting then $Y$ be a random variable taking values $-a,0,b$ with probabilities $p,1-p-r,r$ respectively, we see that $-a\le Y\le b$, $EY=0$, $EY^2=s^2$, and 
$$P(Y>0)=\frac{s^2}{ab+b^2}.$$
So, the lower bound on $P(Y>0)$ in (1) is attained. 

Without the condition $EY^2\ge s^2$, no nonzero lower bound on $P(Y>0)$ exists
even if we still assume that $-a\le Y\le b$ for some real $a\ge0$ and $b\ge0$ -- just let $Y$ be the constant $0$. 
Also, obviously, the exact lower bound $\frac{s^2}{ab+b^2}$ on $P(Y>0)$ goes to $0$ if either $a\to\infty$ or $b\to\infty$. It follows that no nonzero lower bound on $P(Y>0)$ exists if we replace $a$ or $b$ by $\infty$. 
Thus, none of the conditions imposed on $Y$ can be removed if one wants to have a nonzero lower bound on $P(Y>0)$. 
A: The Cantelli inequality asserts that
$$
\Pr(X-\mathbb{E}[X]\ge\lambda)\quad\begin{cases}
\le \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if } \lambda > 0, \\[8pt]
\ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if }\lambda < 0
\end{cases}
$$
for square integrable $X$ with $\sigma^2$ its variance.
A: Not sure how interesting it is, given that computing $\mathbb{E}[|X-\mathbb{E}[X]|]$ may be unwiedly, but Iosif Pinelis' argument can be adapted to give the following statement, which does not require existence of a finite second moment nor a lower bound on the support.
Suppose $Y := X - \mathbb{E}[X]$ satisfies $Y \leq a$ a.s., for some $a>0$. Then
$$
\mathbb{P}\{ Y > 0\} \geq \frac{\mathbb{E}[|Y|]}{2a}\,.
$$
Note that this is achieved for, e.g., $Y$ Rademacher; and that it improves on the variance-base bound from Iosif Pinelis' answer in some cases. (For instance, $Y$ uniform on $[-1,1]$, where we get $1/2$ instead of $1/6$ as a lower bound.)
The proof is just adapting Iosif's, by writing
$$
\mathbf{1}_{Y>0} \geq \frac{Y+|Y|}{2a}
$$
and taking expectations.
