prove a challenging inequality or find a counterexample to it

Suppose $\mathcal{M}_1$ represents the space of smooth probability density functions with unit mean, whose support is contained in $[0,\infty)$ (or $\mathbb{R}_+$). Define the following functional $$\mathrm{J}(f):= \int_0^\infty x\frac{(f')^2}{f} \mathrm{d}x$$ for $f \in \mathcal{M}_1.$ I am conjecturing that $$\mathrm{J}(\rho) \leq \mathrm{J}(f) \quad \text{with} \quad \rho(x): = \int_{z\geq x} \frac{(f*f)(z)}{z} \mathrm{d}z,$$ in which $f*f$ denotes the self-convolution of $f$. I have tried some specific examples (even though not too many) and I did not find any counter-examples (analytically or numerically), so I am wondering whether there exists a proof of this conjecture/inequality, if not, a counter-example is welcome!

Edit: The motivation behind this question is as follows. Take a random variable $X$ with law $f \in \mathcal{M}_1$ unit mean and supported on $[0,\infty)$, we can think of $\mathrm{J}(f)$ as the information contained in $X$ (note that this is not the usual Fisher information, which is often denoted by $\mathrm{I}(f)$ or $\mathrm{I}(X)$). Here $\rho$ is the law of $U(X+Y)$ with $U \sim \mathrm{Uniform}([0,1])$ and $Y$ being an i.i.d. copy of $X$, in which $U$ is also independent of $X$ and $Y$. I am conjecting that the $\mathrm{J}$ information of $U(X+Y)$ is no larger than the $\mathrm{J}$ information of $X$.

Solution 1:

It turns out there is a very simple counter-example (shame on me!). Simply take $f$ to be the uniform distribution $f(x) = \frac 12\mathbf{1}_{[0,2]}(x)$.