Any sols? This one is interesting Progress: Obviously WLOG a>b>c means that we need bc>a. So given any 6, our minimum contradiction would be 1, 2, 3, 4, 5, and then 19, since 4x5>19, so n should be 19.

Let $a_1 <a_2 <...<a_6$ For $n=108$ and $a_1=1,a_2=2,a_3=3,a_4=6,a_5=18,a_6=108$ we can not find such $3$ integers, so $n<108$ Now we prove that $n=107$ is answer Let for $n=107$ statement is false for some $6$ integers, and so we have $a_{i}a_{i+1} \leq a_{i+2}$ for $i=1,2,3,4$ $a_4 \leq \frac{a_6}{a_5}, a_3 \leq \frac{a_5}{a_4},a_2 \leq \frac{a_4}{a_3}$ so $a_2a_3a_4 \leq \frac{a_6}{a_3} \to a_4 \leq \frac{a_6}{a_2a_3^2} \leq \frac{107}{2*3^2}<6<a_2a_3 \leq a_4$ so statement is true for $n=107$