Selection tests for the Balkan Mathematical Olympiad 2013

Assume, without loss of generality, that $$0\le a\le b\le c\le d\le e$$ Because $$ea\le \max \{eb,cd\}\le ec\le ed$$ to get the smallest possible maximum, the best arrangement around the circle is In this case, the maximum is given by $$\max \{eb,cd\}.$$ We have $$eb\le e\frac{b+c+d}{3}\le \frac{e(1-e)}{3}\le \frac{1}{12},$$ and the equality holds when $a=0$, $b=c=d=\frac{1}{6}$, and $e=\frac{1}{2}$. We have, on the other hand, $$cd=\sqrt[3]{c^2{d}^2(cd)}\le (\sqrt[3]{cde}{)}^2\le {\left(\frac{c+d+e}{3}\right)}^2\le \frac{1}{9}$$ and the equality holds when $a=b=0$, and $c=d=e=\frac{1}{3}$. Therefore, the minimum possible value of $T$ is $\frac{1}{9}$.