[TEX]A + B + C = 180 = \pi[/TEX]
mình làm đc đến khúc: sin(B + C) + sin(A + C) + sin(A + B) rùi 0 biết dồn biến ntn :-ss
Cách 1:
Bổ đề:
[TEX]x,y \in [0;\pi] : sinx+siny \leq 2sin(\frac{x+y}{2})[/TEX]
Thật vậy: ta có:
[TEX]sinx+siny=2sin(\frac{x+y}{2})cos(\frac{x-y}{2})[/TEX]
Do [TEX]x,y \in [0;\pi] \Rightarrow \frac{x+y}{2} \in [0;\pi],\frac{x-y}{2} \in [-\frac{\pi}{2};\frac{\pi}{2}][/TEX]
[TEX]\Rightarrow sin(\frac{x+y}{2}),cos(\frac{x-y}{2}) \geq 0[/TEX]
[TEX]\Rightarrow sinx+siny \leq 2sin(\frac{x+y}{2})( do:cos(\frac{x-y}{2}) \leq 1[/TEX]
\Rightarrow ĐPCM
Áp dụng:
Ta có:
[TEX]sinA+sinB+sinC+sin(\frac{\pi}{3}) \leq 2sin(\frac{A+B}{2})+2sin(\frac{C+\frac{\pi}{3}}{2})=2[sin(\frac{A+B}{2})+sin(\frac{C+\frac{\pi}{3}}{2})][/TEX]
[TEX]\leq 4sin(\frac{A+B+C+\frac{\pi}{3}}{4})=4sin(\frac{\pi}{3})=2\sqrt{3}[/TEX]
[TEX]\Leftrightarrow sinA+sinB+sinC \leq \frac{3\sqrt{3}}{2}[/TEX]
Đẳng thức xảy ra \Leftrightarrow tam giác ABC đều
Cách 2:
Ta có:[TEX]sinC=sin(A+B)=sinAcosB+sinBcosA[/TEX]
[TEX]sinA+sinB+sinC=\frac{2}{\sqrt{3}}[\frac{\sqrt{3}}{2}sinA+\frac{\sqrt{3}}{2}sinB]+\sqrt{3}[\frac{sinA}{\sqrt{3}}+\frac{sinB}{\sqrt{3}}cosA][/TEX]
[TEX]\leq(Cauchy)\frac{1}{\sqrt{3}}[(sin^2A+\frac{3}{4})+(sin^2B+\frac{3}{4})]+\frac{\sqrt{3}}{2}[(\frac{sin^2A}{3}+cos^2B)+(\frac{sin^2B}{3}+cos^2A)]=\frac{3\sqrt{3}}{2}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn