Xác định các góc của \(\Delta ABC\) biết \(\left( {1 + \frac{1}{{\sin A}}} \right)\left( {1 + \frac{1}{{\sin B}}} \right)\left( {1 + \frac{1}{{\sin C}}} \right) \)\(= {\left( {1 + \frac{1}{{\sqrt[3]{{\sin A\sin B\sin C}}}}} \right)^3}\).

\(\begin{array}{l}\left( {1 + \frac{1}{{\sin A}}} \right)\left( {1 + \frac{1}{{\sin B}}} \right)\left( {1 + \frac{1}{{\sin C}}} \right) \\= {\left( {1 + \frac{1}{{\sqrt[3]{{\sin A\sin B\sin C}}}}} \right)^3}\\ \Leftrightarrow \frac{{\left( {\sin A + 1} \right)\left( {\sin B + 1} \right)\left( {\sin C + 1} \right)}}{{\sin A\sin B\sin C}} \\= \frac{{{{\left( {\sqrt[3]{{\sin A\sin B\sin C}} + 1} \right)}^3}}}{{\sin A\sin B\sin C}}\\ \Leftrightarrow \sin A\sin B\sin C\\ + \sin A\sin B + \sin B\sin C + \sin A\sin C \\+ \sin A + \sin B + \sin C + 1\\= \sin A\sin B\sin C \\+ 3\sqrt[3]{{{{\sin }^2}A{{\sin }^2}B{{\sin }^2}C}} \\+ 3\sqrt[3]{{\sin A\sin B\sin C}} + 1\\ \Leftrightarrow \sin A\sin B + \sin B\sin C + \sin A\sin C \\+ \sin A + \sin B + \sin C \\= 3\sqrt[3]{{{{\sin }^2}A{{\sin }^2}B{{\sin }^2}C}} \\+ 3\sqrt[3]{{\sin A\sin B\sin C}}\end{array}\)

Ta có \(A,\,\,B,\,\,C\) là các góc trong tam giác \( \Rightarrow 0 < \sin A,\sin B,\sin C \le 1\)

Áp dụng bất đẳng thức Cô-si ta được:

\(\begin{array}{l}\left\{ \begin{array}{l}\sin A\sin B + \sin B\sin C + \sin A\sin C \\\ge 3\sqrt[3]{{{{\sin }^2}A{{\sin }^2}B{{\sin }^2}C}}\\\sin A + \sin B + \sin C\\ \ge 3\sqrt[3]{{\sin A\sin B\sin C}}\end{array} \right.\\ \Rightarrow \sin A\sin B + \sin B\sin C + \sin A\sin C \\+ \sin A + \sin B + \sin C \\\ge 3\sqrt[3]{{{{\sin }^2}A{{\sin }^2}B{{\sin }^2}C}} \\+ 3\sqrt[3]{{\sin A\sin B\sin C}}\end{array}\)

Dấu “\( = \)” xảy ra \( \Leftrightarrow \sin A = \sin B = \sin C\) mà \(A,\,\,B,\,\,C\)  là các góc trong \(\Delta ABC\)

\( \Rightarrow A = B = C = {60^o}\)