Selection tests for the Balkan Mathematical Olympiad, 2013

Assume, without loss of generality, that
$$0 \leq a \leq b \leq c \leq d \leq e$$
Because
$$e a \leq \max \{e b, c d\} \leq e c \leq e d$$
to get the smallest possible maximum, the best arrangement around the circle is

In this case, the maximum is given by
$$\max \{e b, c d\} .$$
We have
$$e b \leq e \frac{b+c+d}{3} \leq \frac{e(1-e)}{3} \leq \frac{1}{12},$$
and the equality holds when $a=0$, $b=c=d=\frac{1}{6}$, and $e=\frac{1}{2}$.
We have, on the other hand,
$$c d=\sqrt[3]{c^{2} d^{2}(c d)} \leq (\sqrt[3]{c d e})^{2} \leq \left(\frac{c+d+e}{3}\right)^{2} \leq \frac{1}{9}$$
and the equality holds when $a=b=0$, and $c=d=e=\frac{1}{3}$.
Therefore, the minimum possible value of $T$ is $\frac{1}{9}$.