Chủ đề này có 2 trả lời

[vietfrog]

Cho tam giác ABC nhọn. Chứng minh rằng :

$\sin (\dfrac{{\pi - A}}{4})\sin (\dfrac{{\pi - B}}{4})\sin (\dfrac{{\pi - C}}{4}) \ge \sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$

[truclamyentu]

Cho tam giác ABC nhọn. Chứng minh rằng :

$\sin (\dfrac{{\pi - A}}{4})\sin (\dfrac{{\pi - B}}{4})\sin (\dfrac{{\pi - C}}{4}) \ge \sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$

$\begin{array}{l}\sqrt {\sin \dfrac{A}{2}\sin \dfrac{B}{2}} = \sqrt {\dfrac{1}{2}\left( {c{\rm{os}}\left( {\dfrac{{A - B}}{2}} \right) - c{\rm{os}}\left( {\dfrac{{A + B}}{2}} \right)} \right)} \le \sqrt {\dfrac{1}{2}\left( {1 - c{\rm{os}}\left( {\dfrac{{A + B}}{2}} \right)} \right)} \\\\= \sin \left( {\dfrac{{A + B}}{4}} \right) = \sin \left( {\dfrac{{\pi - C}}{4}} \right)\end{array}$

Làm tương tự ta có dpcm

[vietfrog]

$\begin{array}{l}\sqrt {\sin \dfrac{A}{2}\sin \dfrac{B}{2}} = \sqrt {\dfrac{1}{2}\left( {c{\rm{os}}\left( {\dfrac{{A - B}}{2}} \right) - c{\rm{os}}\left( {\dfrac{{A + B}}{2}} \right)} \right)} \le \sqrt {\dfrac{1}{2}\left( {1 - c{\rm{os}}\left( {\dfrac{{A + B}}{2}} \right)} \right)} \\\\= \sin \left( {\dfrac{{A + B}}{4}} \right) = \sin \left( {\dfrac{{\pi - C}}{4}} \right)\end{array}$

Làm tương tự ta có dpcm

Chà..Cảm ơn anh nhiều!