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

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

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 sức cô si ta có:

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

Dấu \( = \) xảy ra \( \Leftrightarrow \sin A = \sin B = \sin C\)

\( \Rightarrow \widehat A = \widehat B = \widehat C = {60^0}\)