Tính các góc của tam giác ABC biết ( 1 + 1 s i n A ) ( 1 + 1 s i n B ) ( 1 + 1 s i n C ) = ( 1 + 1 3 √ s i n A . s i n B . s i n C ) 3

\[\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinA}}}}} \right)\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinB}}}}} \right)\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinC}}}}} \right){\rm{ = }}{\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\sqrt[{\rm{3}}]{{{\rm{sinA}}{\rm{.sinB}}{\rm{.sinC}}}}}}} \right)^{\rm{3}}}\]

\[ \Leftrightarrow \frac{{\left( {sinA + 1} \right)\left( {sinB + 1} \right)\left( {sinC + 1} \right)}}{{sinA.sinB.sinC}} = {\left( {\frac{{\sqrt[3]{{sinA.sinB.sinC}} + 1}}{{\sqrt[3]{{sinA.sinB.sinC}}}}} \right)^3}\]

\[ \Leftrightarrow \left( {sinA + 1} \right)\left( {sinB + 1} \right)\left( {sinC + 1} \right) = {\left( {\sqrt[3]{{sinA.sinB.sinC}} + 1} \right)^3}\]

\[ \Leftrightarrow sinA.sinB.sinC + sinA.sinB + sinB.sinC + \sin A.sinC + \sin A + \sin B + \sin C + 1\]

\[ = sinA.sinB.sinC + 3\sqrt[3]{{sinA.sinB.sinC}} + 3\sqrt[3]{{{{\left( {sinA.sinB.sinC} \right)}^2}}} + 1\]

\[ \Leftrightarrow sinA.sinB + sinB.sinC + \sin A.sinC + \sin A + \sin B + \sin C\]

\[ = 3\sqrt[3]{{sinA.sinB.sinC}} + 3\sqrt[3]{{{{\left( {sinA.sinB.sinC} \right)}^2}}}\]

Ta có A, B, C là các góc trong tam giác \[ \Rightarrow {\rm{0 < sinA, sinB, sinC}} \le 1\]

Áp dụng bất đẳng sức cô si ta có:

\(\left\{ {\begin{array}{*{20}{c}}{sinA.sinB + sinB.sinC + \sin A.sinC \ge 3\sqrt[3]{{si{n^2}A.si{n^2}B.si{n^2}C}}}\\{sinA + sinB + sinC \ge 3\sqrt[3]{{sinA.sinB.sinC}}}\end{array}} \right.\)

\[ \Rightarrow sinA.sinB + sinB.sinC + sinAsinC + sinA + sinB + sinC\]

\[ \ge 3\sqrt[3]{{si{n^2}A.si{n^2}B.si{n^2}C}} + 3\sqrt[3]{{sinA.sinB.sinC}}\]

Dấu = xảy ra \[ \Leftrightarrow {\rm{sinA = sinB = sinC}}\]

\[\widehat A{\rm{ = }}\widehat B{\rm{ = }}\widehat C{\rm{ = 6}}{{\rm{0}}^{\rm{0}}}\]

Đáp án cần chọn là: A