Does there exist $c>0$ such that among any $n$ positive integers one may find $3$ with least common multiple at least $cn^3$?
UPDATE
Let me post here a proof that we may always find two numbers with lcm at least $cn^2$. Note that if $a < b$, $N$=lcm$(a,b)$, then $N(b-a)$ is divisible by $ab$, hence $N\geq ab/(b-a)$. So, it suffices to find $a$, $b$ such that $ab/(b-a)\geq cn^2$, or $1/a-1/b\geq 1/a-1/b\leq c^{-1} n^{-2}$. Since at least $n/2$ our numbers are not less then $n/2$, denote them $n/2\leq a_1 < a_2 < \dots < a_k$, $$2/n\geq \sum (1/a_i-1/a_{i+1})\geq k \min (1/a_i-1/a_{i+1}),$$ so $\min (1/a-1/b)\leq 2/nk\leq 4/n^2$.
For triples we get lower bound about $c(n/\log n)^3$ on this way. Again consider only numbers not less then $n/2$. If all lcm's are less then $c(n/\log n)^3$, then all number itselves are less then some $n^3/2$, so for $2^k < n^3 < 2^{k+1}$ all of them do not exceed $2^k$, hence at least $n/2k$ of them belong to $[2^r,2^{r+1}]$ for the same $r$, then there exist three numbers $a < b < c$ with $c-a\leq 2^r/(n/4k)\leq 4ka/n$. Then
lcm$(a,b,c)\geq abc/((b-a)(c-a)(c-b))\geq (a/(c-a))^3\geq (n/4k)^3$.
