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[bachhammer]

Chứng minh rằng với mọi tam giác ABC ta luôn có:
$sinA+sinB+\sqrt{6}sinC\leq \frac{5\sqrt{10}}{4}$.

[Phạm Hữu Bảo Chung]

Giải

Ta có:

$P = \sin{A} + \sin{B} + \sqrt{6}\sin{C} = 2\sin{\dfrac{A + B}{2}}\cos{\dfrac{A - B}{2}} + 2\sqrt{6}\sin{\dfrac{C}{2}}\cos{\dfrac{C}{2}}$

 

$P = 2\cos{\dfrac{C}{2}\cos{\dfrac{A - B}{2}}} + 2\sqrt{6}\sin{\dfrac{C}{2}}\cos{\dfrac{C}{2}}$

 

Do $\cos{\dfrac{C}{2}} > 0 \Rightarrow P \leq 2\cos{\dfrac{C}{2}} + 2\sqrt{6}\sin{\dfrac{C}{2}}\cos{\dfrac{C}{2}}$

 

Mặt khác, ta có: 

$2\cos{\dfrac{C}{2}}\left ( 1 + \sqrt{6}\sin{\dfrac{C}{2}}\right ) = \dfrac{2}{\sqrt{10}}.\sqrt{10}\cos{\dfrac{C}{2}}\left ( 1 + \sqrt{6}\sin{\dfrac{C}{2}}\right )$

 

$\leq \dfrac{1}{\sqrt{10}} \left ( 10\cos^2{\dfrac{C}{2}} + 1 + 2\sqrt{6}\sin{\dfrac{C}{2}} + 6\sin^2{\dfrac{C}{2}}\right )$ 

 

$= \dfrac{1}{\sqrt{10}}\left [ - \left ( 2\sin{\dfrac{C}{2}} - \dfrac{\sqrt{6}}{2}\right)^2 + \dfrac{50}{4} \right ]\leq \dfrac{5\sqrt{10}}{4}$

 

Vậy: $\sin{A} + \sin{B} + \sqrt{6}\sin{C} \leq \dfrac{5\sqrt{10}}{4}$

 

Dấu "=" xảy ra khi: $\left\{\begin{matrix}A = B \\\sin{\dfrac{C}{2}} = \dfrac{\sqrt{6}}{4}\\\cos{\dfrac{C}{2}} = \dfrac{\sqrt{10}}{4}\end{matrix}\right.$

 

(Sử dụng được BĐT Côsi do $0 < C < 180^o$ nên $\sin{\dfrac{C}{2}}, \cos{\dfrac{C}{2}} > 0$ )

Edited by Phạm Hữu Bảo Chung, 15-08-2013 - 15:41.