TITLE: Expected Value Bounded by Median - a Direct Proof?
QUESTION [6 upvotes]: Let $f:[0,1]\to \mathbb{R}$ be a positive, continuous and strictly increasing function such that
$$\int_0^1 f(x)dx=1$$
Let $\xi\in (0,1)$ be such that
$$\int_0^{\xi}f(x)dx=\int_{\xi}^1f(x)dx=\frac{1}{2}$$
Prove that
$$\int_0^1 xf(x)dx\leq \xi$$
This problem comes from physics and has to do with the center of gravity of a shape. But I'd like to see a direct, computational proof. Here is my attempt. To begin with, note that
$$
\begin{aligned}
\int_0^1xf(x)dx
&=\int_0^1[\xi+(x-\xi)]f(x)dx \\
&=\xi\int_0^1f(x)dx+\int_0^1(x-\xi)f(x)dx \\
&=\xi+\int_{-\xi}^{1-\xi}yf(y+\xi)dy
\end{aligned}
$$
For the second term, we know that
$$\int_{-\xi}^{1-\xi}yf(y+\xi)dy=\int_{-\xi}^0yf(y+\xi)dy+\int_0^{1-\xi}yf(y+\xi)dy:=I_1+I_2$$
In $I_1$, let $z=-y$. Since $f$ is strictly increasing, the defining properties of $\xi$ imply that $\xi>\frac{1}{2}$. So $1-\xi<\xi$, and hence
$$
\begin{aligned}
I_1
&=\int_{\xi}^0(-z)(-1)f(-z+\xi)dz \\
&=-\int_0^{\xi}zf(-z+\xi)dz \\
&=-\left[\int_0^{1-\xi}zf(-z+\xi)dz+\int_{1-\xi}^{\xi}zf(-z+\xi)dz\right]
\end{aligned}
$$
We want to show that $I_1+I_2\leq 0$. Equivalently, we have to prove that
$$\int_0^{1-\xi}z\left[f(z+\xi)-f(-z+\xi)\right]dz\leq \int_{1-\xi}^{\xi}zf(-z+\xi)dz$$
If we perform a change of variables on the RHS by setting
$$w=\frac{1-\xi}{2\xi-1}z-\frac{(1-\xi)^2}{2\xi-1}$$
then the RHS becomes
$$\int_0^{1-\xi}\left[(2\xi-1)+\left(\frac{2\xi-1}{1-\xi}\right)^2w\right]f\left((2\xi-1)-\frac{2\xi-1}{1-\xi}w\right)dw$$
However, this is getting too complicated and is clearly not the right way to go. Can anyone share some ideas on how to solve to problem? Thank you!

REPLY [6 votes]: Define $F(x)=\int_0^x f(x)$ and note that is is convex because $f$ is increasing.
Let $\mu=\int_0^1 xf(x)dx$ denote the average. By Jensen's inequality,
$F(\mu)\leq \int_0^1 F(x)f(x)dx$
On the other hand, $\int_0^1 F(x)f(x)dx=\int_0^1 (F(x)^2/2)'dx=1/2$ so we are done.