Let $\xi$ be a nonnegative random variable, and $a>0$. Find $\sup P(\xi\geq a)$ over all distributions such that:
(i): $E(\xi)=20$;
(ii): $E(\xi)=20$, $\text{Var}(\xi)=25$;
(iii): $E(\xi)=20$, $\text{Var}(\xi)=25$ and $\xi$ is symmetric about its mean value.
I can't figure out any one. Here is my thinking for (i): a basic tail bound gives $P(\xi \geq a)\leq \frac{E(\xi)}{a}$, and I want to find a sequence of random variables $(\xi_n)$ satisfying the above constraints and so that $\lim\limits_{n\to\infty}P(\xi_n \geq a) = \frac{20}{a}$.
If $\int_a^\infty f_\xi(x)d x$ were equal to $\frac{20}{a}$ (and differentiable wrt $a$), then $f_\xi(x) = \frac{20}{a^2}$, so I try a sequence of power-law distributions. The one above doesn't have a finite expectation, so I index them by $\varepsilon>0$. Supposing $f_{\xi,\varepsilon}(x)=I_{[k,\infty)}\frac{c}{x^{2+\varepsilon}}$, we have two constraints (integrate to unity and the expectation constraint from the question) and two unknowns, which I solved to find $k=\frac{20\varepsilon}{1+\varepsilon}$ and $c=\frac{(20\varepsilon)^{1+\varepsilon}}{(1+\varepsilon)^\varepsilon}$. For any $\varepsilon>0$ these satisfy the constraint and this is a valid distribution, but $P(\xi_n\geq a)=(\frac{20\varepsilon}{1+\varepsilon})^{1+\varepsilon}\frac{1}{a^{1+\varepsilon}}\not\to\frac{20}{a}$, but to $0$ instead, as $\varepsilon\to0$.
Question: is this idea salvageable, or is there a similar approach that works? Did I just make some silly mistake somewhere? If not, can anyone give some ideas?
I also considered looking at when the first moment inequality above achieves equality, which it does for random variables that depend on $a$, which doesn't seem like it is supposed to be used.
Thanks.