I was solving one of the Number theory problems from Mathematical Olympiad Challenges,
And the problem goes like :

Prove that there is a perfect cube between $n$ and $3n$ for any integer $n\geq 10$.

And experimenting I found that if $a^3$ lies b/w, $n$ and $3n$ ($n≥19$) then it also lies b/w $m$ and $3m$ , where $m<a^3$ ;
Therefore if the statement to be proved is true then taking $m=a^3$ implies b/w $m$ and $3m$ there lies $(a+1)^3$ i.e., $(a+1)^3$ lies b/w $a^3$ and $3a^3$ ,
So therefore to prove this we need to prove an equivalent inequality i.e.,

$n^3 <(n+1)^3 < 3n^3$ ; for any integer $n\geq10$