Let $X$ be a non-negative random variable bounded by $X\le C^2$ and mean $\mathbb{E} X = C$ for some constant $C>1$. Moreover, we know that its PDF is monotonically decreasing in the $[0, C^2]$. Can we upper bound $P(X>\varepsilon C)$, with a quantity that only depends on $\epsilon$?
A trival bound by Markov is $$P(X>\varepsilon C)\le \frac{E X}{\epsilon C} = \frac{1}{\varepsilon}$$ However, because the variable is bounded by $C^2$, for $\varepsilon\ge C$ we have $P(X>\varepsilon C)=0$. Supposing this value is a continuous w.r.t. $\varepsilon$, it is interesting how does the $1/\varepsilon$ bound switches to to $0$ over the range $[0,C]$.
it is pretty clear that in general settings the worst-case behavior would happen when the distribution of $X$ is bimodal, i.e., a Dirac delta at $X=0$ and another Dirac delta $X=C^2$ with values that satisfy $E X = C$. But because of monotonicity, the variable cannot be bimodal. What is the worst-case scenario in this case?
I think you need $C \ge 2$ for this to work at all with the requirement for a non-negative random variable with a monotonically decreasing density (i.e. a mode at $0$) bounded above by $C^2$ and with a mean of $C$. Then I think you have
$\mathbb P(X>\varepsilon C) \le \left\{\begin{array}{lrr} 1, & \text{for }& \varepsilon \leq 0\\ 1- \dfrac\varepsilon 2, & \text{for }& 0 \le \varepsilon \leq 1 \\ \dfrac{1}{2\varepsilon}, & \text{for }& 1 \le \varepsilon \leq \frac C2 \\ \dfrac{2(C-\varepsilon)}{C^2}, & \text{for }& \frac C2 \le \varepsilon \leq C \\ 0, & \text{for }& C \le \varepsilon \qquad \end{array}\right.$
and that this is tight if you consider a distribution which for some $0 \lt p \le 1$ has a cumulative distribution function $\mathbb P(X \le x)= (1-p) + x\frac{p^2}{2C}$ when $x \in \left(0,\frac{2C}{p}\right]$. The optimal values of $p$ seem to be to use $p=1$ when $\varepsilon \leq 1$, $p=\frac{1}{\varepsilon}$ when $1 \le \varepsilon \leq \frac C2$, and $p=\frac{2}{C}$ when $\frac C2 \le \varepsilon $.