Lower bound for (function of) density of well-behaved random variable Suppose we have a non-negative random variable $\tilde{\theta}$ such that $\mathbb{E}\tilde{\theta} = a > 0$, with finite variance $\sigma^2$. $\tilde{\theta}$ can take on values from $0$ to $\overline{\theta} > a$, where $\overline{\theta}$ may be infinite.
Let the random variable's CDF be given by $F(\theta) := \Pr\{\tilde{\theta} \leq \theta\}$, with a strictly positive density function $f(\theta) > 0$ in the interior of its support $[0, \overline{\theta}]$. For simplicity, we can assume that $f(\theta)$ is continuous and differentiable everywhere. 
As an example, $\tilde{\theta}$ could be drawn from a log-normal distribution.
My question is as follows: Given the outlined setup, can we devise a lower bound for $\max_{\theta} \theta f(\theta)$ of the random variable $\tilde{\theta}$?
So far, I only have a partial approach: If the variance of $\tilde{\theta}$ is sufficiently small, we can employ Chebyshev's inequality, which tells us that if a random variable with mean $a$ has a finite variance $\sigma^2$, at least $1 - \frac{1}{k^2}$ of its probability mass must fall in the range $[a - k \sigma, a + k \sigma]$, for $k \geq 1$. 
Hence, for $\sigma < a$, there are values of $k > 1$ that give us a non-trivial lower bound for the probability mass in an interval around $a$ that doesn't encompass zero.
Namely, since for $k > 1$, at least $1 - \frac{1}{k^2}$ of the probability mass of $\tilde{\theta}$ falls in the range $[a - k \sigma, a + k \sigma]$, the average density in this range must be at least $\frac{1 - \frac{1}{k^2}}{2k\sigma}$.
Therefore, for $k \in (1, \frac{a}{\sigma})$, we can deduce that
$max_{\theta}\ \theta f(\theta) \geq (a-k\sigma)\left(\frac{1 - \frac{1}{k^2}}{2k\sigma}\right)$.
My question is now: Can we do more? Most importantly, can we say something for an arbitrary variance $\sigma^2$, in particular if $\overline{\theta}$ is infinite?
Many thanks in advance!
 A: Think of it as a calculus-of-variations problem:
$$ \eqalign{\text{minimize}\ &r\cr
\text{subject to}\ & \int_0^\infty f(\theta)\; d\theta = 1\cr
                 &\int_0^\infty \theta f(\theta)\; d\theta = a \cr
                 &\int_0^\infty \theta^2 f(\theta)\; d\theta = \sigma^2 + a^2\cr
& 0 \le \theta f(\theta) \le r,\ 0 < \theta < \infty\cr}$$
The trial functions to consider are of the form 
$$ f_{bcr}(\theta) = \cases{r/\theta & for $b \le \theta \le c$\cr
0 & otherwise\cr}$$
This satisfies the constraints if
$$ \eqalign{ r \ln(c/b) &= 1\cr
             r (c-b) &= a\cr
             r (c^2-b^2)/2 &= \sigma^2 + a^2\cr}$$
The first two equations can be solved for $b$ and $c$:
$$ \eqalign{
            b &= \dfrac{a}{r(e^{1/r}-1)}\cr
            c &= \dfrac{a\; e^{1/r}}{r(e^{1/r}-1)}\cr}$$
and then from the third equation 
$$ \dfrac{e^{1/r}+1}{r(e^{1/r}-1)} = 2 + 2 \dfrac{\sigma^2}{a^2} 
$$
That does not have a closed-form solution for $r$, but here is its
graph as a function of $\sigma^2/a^2$.

A: If $\mathbb{E}\tilde{\theta}=a$, then $\max\theta f(\theta)\ge a$.
Consider $\tilde{\theta}$ distributed in the interval $(x,y)$ with density $f(\theta)=1/(\theta\log(y/x))$ for $x<\theta<y$ and zero otherwise. This distribution fulfills $\max \theta f(\theta)=\mathbb{E}\tilde{\theta}$. So in general you cannot improve the previous lower bound. One may think that if the random variable has finite variance then $f(\theta)$ would be larger and that may improve the bound. But i does not. For this distribution you can make the variance vanishingly small by taking $x$ close to $y$ or very large by taking them very far apart and you will always have $\max \theta f(\theta)=\mathbb{E}\tilde{\theta}$.
