Find the least natural number $n$, such that the following inequality holds:$\sqrt{\dfrac{n-2011}{2012}}-\sqrt{\dfrac{n-2012}{2011}}<\sqrt[3]{\dfrac{n-2013}{2011}}-\sqrt[3]{\dfrac{n-2011}{2013}}$.
Let $s = n-2011$. we can rewrite the inequation as:
$ \sqrt{\frac{s}{2012}}-\sqrt{\frac{s-1}{2011}}<\sqrt[3]{\frac{s-2}{2011}}-\sqrt[3]{\frac{s}{2013}} $
for $s<2013$: we have
$ \sqrt{\frac{s}{2012}}-\sqrt{\frac{s-1}{2011}}\geq 0>\sqrt[3]{\frac{s-2}{2011}}-\sqrt[3]{\frac{s}{2013}} $
for $s = 2013$: the inequality is obviously true.
Conclusion: the least natural number is $s = 2013$. so: $n=s+2011=4024$