Đặt \(\left(\frac{1}{sinA};\frac{1}{sinB};\frac{1}{sinC}\right)=\left(a;b;c\right)\Rightarrow a;b;c>0\), áp dụng BĐT AM-GM
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{3}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge\frac{3\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cộng vế với vế và rút gọn: \(1\ge\frac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
\(\Leftrightarrow\left(1+\frac{1}{sinA}\right)\left(1+\frac{1}{sinB}\right)\left(1+\frac{1}{sinC}\right)\ge\left(1+\frac{1}{\sqrt[3]{sinA.sinB.sinC}}\right)^3\)
Dấu "=" xảy ra khi và chỉ khi \(\frac{1}{sinA}=\frac{1}{sinB}=\frac{1}{sinC}\Leftrightarrow\)
\(A=B=C=60^0\)
