Đáp án:
$\begin{array}{l}
{u_n} = n + \frac{{2013}}{{{n^3}}}\\
= \frac{1}{3}n + \frac{1}{3}n + \frac{1}{3}n + \frac{{2013}}{{{n^3}}}\\
\ge 4\sqrt[4]{{\frac{1}{3}n.\frac{1}{3}n.\frac{1}{3}n.\frac{{2013}}{{{n^3}}}}} = 4.\sqrt[4]{{\frac{{671}}{9}}}\\
\Rightarrow GTNN:{u_n} = 4.\sqrt[4]{{\frac{{671}}{9}}} \Leftrightarrow \frac{1}{3}n = \frac{{2013}}{{{n^3}}}\\
\Rightarrow {n^4} = 6039\\
\Rightarrow n = \sqrt[4]{{6039}}
\end{array}$
