China Girls' Mathematical Olympiad

(1) Let $a=b=c=\frac{1}{3}$, $d=e=0$, it is easy to see that, when arranging them around a circle, we can always get two neighboring numbers of $\frac{1}{3}$, and their product is $\frac{1}{9}$.

(2) For any five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$, without loss of generality, we assume that $a\ge b\ge c\ge d\ge e\ge 0$. Arrange these numbers around a circle in such a way as seen in the figure:



Since $a+b+c+d+e=1$, we have $a+3d\le 1$, and $$a\cdot 3d\le {\left(\frac{a+3d}{2}\right)}^2\le \frac{1}{4}$$ then $ad\le \frac{1}{12}$.

Furthermore, $a+b+c\le 1$, then $b+c\le 1-a\le 1-\frac{b+c}{2}$, i.e. $b+c\le \frac{2}{3}$. So $$bc\le \frac{(b+c{)}^2}{4}\le \frac{1}{9}$$ Since $ce\le ae\le ad$ and $bd\le bc$, then any neighboring numbers in this arrangement have their product less than $\frac{1}{9}$.