The problem is stated as follows:
Prove or disprove: there exists some absolute constant $c>0$, satisfying the following statement:
Given any $n$ distinct positive integers $a_i$, $1\leq i\leq n$, then there must exist three integers $1\leq i,j,k\leq n$ such that $[a_i,a_j,a_k]\ge cn^3$.
This problem comes from the two numbers version.
Arrange the number in increasing order, $a_1< a_2< \cdots < a_n$
$$[a_i,a_{i-1}]=\frac{a_ia_{i-1}}{(a_i,a_{i-1})}\ge\frac{a_ia_{i-1}}{a_i-a_{i-1}}$$
Thus if $$\frac{1}{\frac{n}{2}}\ge\frac{1}{a_{\frac{n}{2}}}-\frac{1}{a_n}=\sum_{i=\frac{n}{2}}^{n-1}(\frac{1}{a_i}-\frac{1}{a_{i+1}})\ge\sum_{i=\frac{n}{2}}^{n-1}\frac{1}{[a_i,a_{i+1}]}$$.
Thus there must exist some $i$, so that $[a_i,a_{i+1}]\ge \frac{n^2}{8}$.
However, this method doesn't work for larger than three case. I've known about another proof that can do for $3-\epsilon$.
Assume the contrary, assume that the $n$ positive integers with lcm all in $O(n^{3-\epsilon})$. Then we fix a number t, then the number of tuples $(x,y,z),s.t. t=[x,y,z]$ is obviously at most $\tau(t)^3$. Now we sum over $1\leq t\leq O(n^{3-\epsilon})$, then any $a_i,a_j,a_k$ must appear in the summation. Thus,
$$\sum_{t=1}^{O(n^{3-\epsilon})}\tau(t)^3\ge n^3$$ Use $\tau(t)\leq t^{\frac{\epsilon}{4}}$ when $t$ is large enough gives a contradiction.
So I wonder whether the bound is $O(n^3)$ or not. Can somebody give some references on this problem? Thanks so much!