Prove that there is a perfect cube between n and 3n for any integer n≥10 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 $n\geq10$

Let us assume $n < k^3 < 3n$
This is true for $n=10, k = 3$.
Now by construction, we know there is cube in $k^3 - 1 < k^3 < 3k^3 - 3$
So we are covered for all $n$ from $n$ to $k^3 - 1$.
Now we just have to prove that $k^3 < (k+1)^3 < 3k^3$
Obviously $k^3 < (k+1)^3$ and the second inequality is true for $k \ge 3$
And we can continue this forever.
