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 $P(Y>0)$.

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