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 \(\frac{1}{sinA}=\frac{1}{sinB}=\frac{1}{sinC}\Leftrightarrow\)

\(A=B=C=60^0\)

Lời giải:

Thay dấu "=" thành $\geq $ ta được BĐT Holder. Dấu "=" xác định tại $\sin A=\sin B=\sin C$ hay tam giác $ABC$ đều.

Chứng minh cụ thể như sau:

\(\frac{1}{1+\frac{1}{\sin A}}+\frac{1}{1+\frac{1}{\sin B}}+\frac{1}{1+\frac{1}{\sin C}}\geq 3\sqrt[3]{\frac{1}{(1+\frac{1}{\sin A})(1+\frac{1}{\sin B})(1+\frac{1}{\sin C})}}\)

\(\frac{\frac{1}{\sin A}}{1+\frac{1}{\sin A}}+\frac{\frac{1}{\sin B}}{1+\frac{1}{\sin B}}+\frac{\frac{1}{\sin C}}{1+\frac{1}{\sin C}}\geq 3\sqrt[3]{\frac{\frac{1}{\sin A\sin B\sin C}}{(1+\frac{1}{\sin A})(1+\frac{1}{\sin B})(1+\frac{1}{\sin C})}}\)

Cộng theo vế và rút gọn:

\(\Rightarrow 3\geq 3\frac{1+\sqrt[3]{\frac{1}{\sin A\sin B\sin C}}}{\sqrt[3]{(1+\frac{1}{\sin A})(1+\frac{1}{\sin B})(1+\frac{1}{\sin C})}}\)

\(\Rightarrow (1+\frac{1}{\sin A})(1+\frac{1}{\sin B})(1+\frac{1}{\sin C})\geq (1+\sqrt[3]{\frac{1}{\sin A\sin B\sin C}})^3\)

Dấu "=" xảy ra (như đề bài) khi \(\sin A=\sin B=\sin C\Rightarrow \angle A=\angle B=\angle C=60^0\)

\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)

\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)

\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)

\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)

\(sin\left(\frac{\pi}{7}\right)H=sin\left(\frac{\pi}{7}\right)cos\left(\frac{2\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\)

\(=\frac{1}{2}\left[sin\left(\frac{3\pi}{7}\right)-sin\left(\frac{\pi}{7}\right)+sin\left(\frac{5\pi}{7}\right)-sin\left(\frac{3\pi}{7}\right)+sin\pi-sin\left(\frac{5\pi}{7}\right)\right]\)

\(=-\frac{1}{2}sin\left(\frac{\pi}{7}\right)\)

\(\Rightarrow H=-\frac{1}{2}\)

\(sinA+sinB+sinC=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{C}{2}\right)cos\left(\frac{C}{2}\right)\)

\(=2cos\frac{C}{2}cos\left(\frac{A-B}{2}\right)+2cos\left(\frac{A+B}{2}\right)cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left[cos\left(\frac{A-B}{2}\right)+cos\left(\frac{A+B}{2}\right)\right]\)

\(=4cos\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}\)