Bump......

ssilwa wrote: Bump...... Sorry what is bump

I think we can use MV method WLOG $a_{1}=max(a_{1},a_{2},...,a_{10})$ $f(a_{1},a_{2},...,a_{10})=a_{1}a_{2}+a_{2}a_{3}+...+a_{10}a_{1}$ We can prove that $f(a_{1},a_{2},...,a_{10})\geq f(a_{1}+a_{10}-1,a_{2},...,a_{9},1)$ Also with $f(a_{1}+a_{10}-1,a_{2},...,a_{9},1)\geq f(a_{1}+a_{10}+a_{9}-1,a_{2},...,1,1)$ and so on... So $f(a_{1},a_{2},...,a_{10})\geq f(a_{1}+a_{2}+...+a_{10}-1,...,1,1)$ So the minimum of $f(a_{1},a_{2},...,a_{10})$ is $1995+1+1+...+1+1995=3998$

Sorry I forgot “$a_{1},a_{2},...,a_{10}$ be distinct natural numbers” But I think we can use same method and get $a_{1},a_{2},...,a_{10}\in {{1,2,...,9,1950}}$ And some of those permutations can give us the minimum value