### This question was originally asked on the Mathematics StackExchange by User smcc

Consider a continuous random variable $V$ with cumulative distribution function $F$ and density function $f$. Suppose that $f$ is compactly supported on $[a,b]\subset\mathbb R_{\geq 0}$. Let $\mu$ denote the expected value of $V$.

Conjecture.Then it holds that \begin{equation}\tag 1\label 1 \mu F(\mu)^2\geq\int_{\mu}^{b}F(x)[1-F(x)] \,\mathrm dx. \end{equation}

*Question.* Is this conjecture true? How can we (dis)-prove it?

*Remark.* In the original question, we had the additional assumptions that $f$ should be differentiable on $]a,b[$ and that $\ln\circ f$ should be concave on $[a,b]$. I don't think that these assumptions are necessary though.

*Notation.* For $x\in[a,b]$, let $$\phi(x):=x F(x)^2-\int_{x}^{b}F(y)\cdot(1-F(y))\,\mathrm dy.$$

Then \eqref{1} is $\phi(\mu)\geq0$.

**Partial results.** (Refer to the original question for the derivations)

- We have $\phi(x)=g(x)-h(x)$, where $$g(x)=\int_{x}^{b}yf(y)\,\mathrm dy+\mu F(x), \qquad \text{and} \qquad h(x)=\int_{x}^{b}y\cdot(2f(y)\cdot F(y))\,\mathrm dy.$$ In particular, $\phi(a)<0$ and $\phi(b)>0$. Additionally, $\phi$ is strictly increasing (as $\phi'>0$ on $]a,b[$) and $g$ achieves its maximum at $x=\mu$.
- Inequality \eqref{1} holds if [$\mu$ is greater or equal than the median of $V$] and $\mu\geq \frac b2$. In particular, if $f$ is flip-symmetrical to $\frac{a+b}2$, then \eqref{1} holds.
- Inequality \eqref{1} holds if [$f$ is increasing on $[a,b]$] and $F(\mu)\geq\frac12$.
- We have $$\phi(\mu)\geq \mu F(\mu)^2-\int_{a}^{\mu}F(x)\,\mathrm dx.$$ However, the right-hand side of the last equation can be negative. For example, for $F(x)=\ln(1+(e-1)\cdot x)$ on $[0,1]$, the right-hand side is $\approx-0.0008$. (Remark: The according density function to my $F$ is not $\ln$-concave. I wasn't able to find a $\ln$-concave counter-example).