Let $X$ be a (discrete) random variable that takes values in $[0,1]$ and such that $\textrm{var}(X)$ is big in some sense like $$\textrm{var}(X)\ge\varepsilon$$ Can the mean $\mathbb{E}(X)$ be bounded from below? The resaoning is that, if $\textrm{var}(X)$ is "big", then $X$ takes many different values, so it takes at least one high value and thus the sum is at least as big as that value.
However, I haven't been able to prove anything yet even if I use Markov's and Chebyshev's inequalities.
Many thanks!
As noted in the comments, for a distribution of variance $\epsilon$ supported on $\{0,1\}$, the mean $p$ solves $$p-p^2=\epsilon \,. \tag{*}$$ Let $$p_1= \frac{1-\sqrt{1-4\epsilon}}2 \in [0, 1/2]$$ be the smaller solution of $(*)$.
Consider a random variable $X$ supported in $[0,1]$, of mean $m$ and variance $\epsilon$. We have $$m^2+\epsilon=E[X^2] \le E[X]=m\,,$$ so $m-m^2 \ge \epsilon $, whence $\epsilon \le 1/4$ and $m \ge p_1$.