Jumbler wrote: Let $n$ be a positive integer with $n \geq 2$, and $0<a_{1}, a_{2},...,a_{n}< 1$. Find the maximum value of the sum $\sum_{i=1}^{n}(a_{i}(1-a_{i+1}))^{\frac{1}{6}}$ where $a_{n+1}=a_{1}$ Not hard:\[\sum_{i=1}^{n}{(a_{i}(1-a_{i+1}))^{\frac{1}{6}}}=2^{\frac32}\sum_{i=1}^{n}(\frac12\frac12\frac12\frac12a_{i}(1-a_{i+1}))^{\frac{1}{6}}\leq \sum_{i=1}^{n}2^{\frac32}\frac{1-a_{i+1}+a_{i}}{6}\]