Let $X$ be a random variable with $X \in [a,b]$, known mean $\mathbb{E}[X] = \mu$ and known variance $\text{var}[X] = \sigma^2$. We are interested in an expression for the worst-case quantile. That is, for $\epsilon \in [0,1]$ we wish to find $R$ such that $$\begin{equation}\label{eq: 1} \sup_{\mathbb{P} \in \mathcal{P}} \mathbb{P} [X < R] = \epsilon, \end{equation}$$ where \begin{equation} \mathcal{P} = \big\{\mathbb{P} : \text{supp}(X) \in [a,b],~\mathbb{E}_{\mathbb{P}}[X] = \mu,~\text{var}[X] = \sigma^2 \big\}, \end{equation} is the set of all possible distribution functions. I have tried an approach using the Markov inequality below. However, I tend to find values of $R$ that are outside of $[a,b]$. Is there a mistake in my reasoning below? In particular, it does not use the lower bound $a$. Or is there a better approach?
The Markov inequality states for a nonnegative random variable $Y$ and a constant $R>0$ that $\mathbb{P}[Y \geq R] \leq \frac{\mathbb{E}[Y]}{R}$. It is easily derived that for a random variable $X$ with $\mathbb{P}[X \leq b] = 1$ and a constant $R \leq \mathbb{E}[X]$ we have $\mathbb{P}[X\leq R] = \frac{b-\mathbb{E}[X]}{b-R}$ (use the Markov inequality for a random variable $\hat{X} = b-X$). Also, the Markov inequality can be shown to be tight. Therefore, we obtain \begin{align} \sup_{\mathbb{P} \in \mathcal{P}} \mathbb{P} [X < R] = \frac{b-\mu}{b-R}, \end{align} and combining this with the first equation gives $R = \frac{\mu}{\epsilon} - \frac{1-\epsilon}{\epsilon}b$. However, this implies that we only have $R \geq a$ if $\epsilon \geq \frac{b-\mu}{b-a}$. For example, with $[a,\mu,b] = \big[\frac{1}{20},\frac{1}{10},\frac{1}{4}\big]$, this gives $\epsilon = 0.75$. For such a bound to be useful, we wish to have that $R \in[a,b]$ for all values of $\epsilon$.