$\begin{array}{l} S = \sum\limits_{cyc}^{a,b,c} {\dfrac{{{a^n}}}{{1 + n{b^{n + 1}}}}} \\ S = \sum\limits_{cyc}^{a,b,c} {\left( {{a^n} - \dfrac{{n{a^n}{b^{n + 1}}}}{{1 + n{b^{n + 1}}}}} \right)} \\ S = \sum\limits_{sum}^{a,b,c} {{a^n} - \sum\limits_{cyc}^{a,b,c} {\dfrac{{n{a^n}{b^{n + 1}}}}{{1 + n{b^{n + 1}}}}} } \\ S \ge k - \sum\limits_{cyc}^{a,b,c} {\dfrac{{n{a^n}{b^{n + 1}}}}{{\left( {n + 1} \right)b}} = k - } \dfrac{n}{{n + 1}}\sum\limits_{cyc}^{a,b,c} {{a^n}{b^n}} \end{array}$
Mặt khác ta có:
$\sum\limits_{cyc}^{a,b,c} {{a^n}{b^n} = {a^n}{b^n} + {b^n}{c^n} + {c^n}{a^n}} \le \dfrac{{{{\left( {{a^n} + {b^n} + {c^n}} \right)}^2}}}{3} = \dfrac{{{k^2}}}{3}$
Suy ra $S \ge k - \dfrac{n}{{n + 1}}.\dfrac{{{k^2}}}{3}$
Dấu bằng xảy ra khi và chỉ khi $a = b = c = \dfrac{k}{3}$
