I think I have managed to show an inequality about probability density functions and their expected values, but I can't seem to find any literature mentioning it. Maybe it takes another form in another context? I'm not sure what to do with it other than ask if anyone knows about it or maybe have a way to show that it is trivial (or even wrong).
The result goes:
Let $X$ be a continuous stochastic variable with range $R_X=[a,b] \subseteq \mathbb{R}$ and PDF $f_X(x)$. Also let $c \in R_X$ be a constant. Then the following ineqalities apply for all $c$ in $R_X$: \begin{align} \int_c^b f_X(x)(c-x)dx \leq \frac{c-E[X]}{2} \leq \int_a^c f_X(x)(c-x)dx. \end{align}
Proof. Construct the following two inequalities from the fact that $x \leq c$ for $x \in [a,c]$ and $c \leq x$ for $x \in [c,b]$: \begin{align} \int_a^c xf_X(x)dx \leq \int_a^c cf_X(x)dx, \qquad \int_c^b cf_X(x)dx \leq \int_c^b xf_X(x)dx. \tag{1} \end{align}
Add each side of the one inequality to the corresponding side of the the other: \begin{align} \int_a^c xf_X(x)dx+\int_c^b cf_X(x)dx \leq \int_a^c cf_X(x)dx+\int_c^b xf_X(x)dx. \tag{2} \end{align}
Subtract such that the limits of integration are the same on each side: \begin{align} \int_a^c xf_X(x)dx-\int_a^c cf_X(x)dx \leq \int_c^b xf_X(x)dx-\int_c^b cf_X(x)dx. \tag{3} \end{align}
\begin{align} \int_a^c f_X(x)(x-c)dx \leq \int_c^b f_X(x)(x-c)dx. \tag{4} \end{align}
Add $\int_a^c f_X(x)(x-c)dx$ to both sides: \begin{align*} \int_a^c f_X(x)(x-c)dx+\int_a^c f_X(x)(x-c)dx \leq \int_a^c f_X(x)(x-c)dx+\int_c^b f_X(x)(x-c)dx. \tag{5} \end{align*}
\begin{align} 2\int_a^c f_X(x)(x-c)dx &\leq \int_a^b f_X(x)(x-c)dx \\\\ &= \int_a^b xf_X(x)dx-c\int_a^b f_X(x)dx \\\\ &= E[X]-c. \tag{6} \end{align}
By rearranging, we get the second part of the double inequality: \begin{align} \frac{c-E[X]}{2} \leq \int_a^c f_X(x)(c-x)dx. \tag{7} \end{align}
The first part of the double inequality is obtained by adding $\int_c^b f_X(x)(x-c)dx$ on both sides in $(5)$ and proceeding the same way. Thus we obtain: \begin{align} \int_c^b f_X(x)(c-x)dx \leq \frac{c-E[X]}{2} \leq \int_a^c f_X(x)(c-x)dx \tag{8} \end{align} as was to be shown.
The integals slightly resembles a convolution for certain integration limits, but I am not sure if that would be a sensical generalization. Would love some insights. Thanks in advance!
The inequalities can be re-written as $$ \mathbb E[(c-X) \mathbf 1_{X>c}] \le \frac{\mathbb E[c-X]}{2} \le \mathbb E[(c-X) \mathbf 1_{X < c}]. $$ Let $Y = c - X$, thus the range of $Y$ must contain the origin. Then, $$ \mathbb E[Y \mathbf 1_{Y<0}] \le \frac{\mathbb E[Y]}{2} \le \mathbb E[Y \mathbf 1_{Y > 0}]. $$
The upper bound is interesting only if $\mathbb E[Y] \ge 0$, but then we have a better upper bound, namely $$\mathbb E[Y] \le E[Y \mathbf 1_{Y > 0}].$$
Analogously, the lower bound is interesting only if $\mathbb E[Y] \le 0$, but then we have a better lower bound, namely $$\mathbb E[Y] \ge E[Y \mathbf 1_{Y < 0}].$$