If the numbers are $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$, $0$, $0$, then there will surely be two neighbours equal to $\frac{1}{3}$, with product $\frac{1}{9}$. Conversely, we will show an ordering under which the considered products do not exceed $\frac{1}{9}$. If the numbers are $0 \leq a \leq b \leq c \leq d \leq e$, then let the ordering be $e$, $a$, $d$, $c$, $b$, with products $ae$, $ad$, $cd$, $bc$ and $be$. Since $ad \leq ae \leq be$ and $bc \leq cd$, it is enough to argue $be \leq \frac{1}{9}$ and $cd \leq \frac{1}{9}$. If we suppose that $c+d > \frac{2}{3}$, then we would obtain $d + e > c + e > c + d > \frac{2}{3}$, so $c+d+e > 1$, contradiction. Therefore $c + d \leq \frac{2}{3}$ and so $cd \leq \frac{(c+d)^2}{4} \leq \frac{1}{9}$. If we suppose that $be > \frac{1}{9}$, then clearly $d \geq c \geq b > \frac{1}{9e}$, whence $1 = a+b+c+d+e > e + \frac{1}{3e} \geq \frac{2}{\sqrt{3}}$ (equivalent to $\left(e - \frac{1}{\sqrt{3}}\right)^2 \geq 0$), contradiction. Therefore $be \leq \frac{1}{9}$. (Actually, in the same manner it follows that $be \leq \frac{1}{12}$.)