Tương đương: ${\sin ^{2m}}x.{\cos ^{2n}}x \le \dfrac{{{m^m}{n^n}}}{{{{\left( {m + n} \right)}^{m + n}}}}$
Đặt ${\sin ^2}x = a \in \left[ { - 1;1} \right]$
Ta có $f\left( a \right) = {a^m}{\left( {1 - a} \right)^n}$
$f'\left( a \right) = m{a^{m - 1}}{\left( {1 - a} \right)^n} - n{a^m}{\left( {1 - a} \right)^{n - 1}}$
Với $a \in \left[ {\dfrac{m}{{m + n}};1} \right] \Rightarrow f'\left( a \right) \le 0 \Rightarrow f\left( a \right) \le f\left( {\dfrac{m}{{m + n}}} \right) = \dfrac{{{m^m}{n^n}}}{{{{\left( {m + n} \right)}^{m + n}}}}$
Với $a \in \left[ { - 1;\dfrac{m}{{m + n}}} \right) \Rightarrow f'\left( a \right) \ge 0 \Rightarrow f\left( a \right) \le f\left( {\dfrac{m}{{m + n}}} \right) = \dfrac{{{m^m}{n^n}}}{{{{\left( {m + n} \right)}^{m + n}}}}$
Tóm lại ta luôn có: ${\sin ^m}x.{\cos ^n}x \le \sqrt {\dfrac{{{m^m}{n^n}}}{{{{\left( {m + n} \right)}^{m + n}}}}} $ Q.E.D