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Cách 1:
Bổ đề:
$x,y \in [0;\pi] : sinx+siny \le 2sin(\dfrac{x+y}{2})$
Thật vậy: ta có:
$sinx+siny=2sin(\dfrac{x+y}{2})cos(\dfrac{x-y}{2})$
Do $x,y \in [0;\pi] \Longrightarrow \dfrac{x+y}{2} \in [0;\pi],\dfrac{x-y}{2} \in [-\dfrac{\pi}{2};\dfrac{\pi}{2}]$
\Rightarrow $sin(\dfrac{x+y}{2}),cos(\dfrac{x-y}{2}) \ge 0$
\Rightarrow $sinx+siny \le 2sin(\dfrac{x+y}{2})$ ( do:$cos(\dfrac{x-y}{2}) \le 1$
\Rightarrow ĐPCM
Áp dụng:
Ta có:
$sinA+sinB+sinC+sin(\dfrac{\pi}{3}) \le 2sin(\dfrac{A+B}{2})+2sin(\dfrac{C+\dfrac{\pi}{3}}{2})=2[sin(\frac{A+B}{2})+sin(\frac{C+\frac{\pi}{3}}{2})]$
$\le 4sin(\dfrac{A+B+C+\dfrac{\pi}{3}}{4})=4sin(\dfrac{\pi}{3})=2\sqrt{3}$
\Leftrightarrow $sinA+sinB+sinC \le \dfrac{3\sqrt{3}}{2}$
Đẳng thức xảy ra \Leftrightarrow tam giác ABC đều
Cách 2:
Ta có:$sinC=sin(A+B)=sinAcosB+sinBcosA$
$sinA+sinB+sinC=\dfrac{2}{\sqrt{3}}[\dfrac{\sqrt{3}}{2}sinA+\dfrac{\sqrt{3}}{2}sinB]+\sqrt{3}[\dfrac{sinA}{\sqrt{3}}+\dfrac{sinB}{\sqrt{3}}cosA]$
$\le (Cauchy) \dfrac{1}{\sqrt{3}}[(sin^2A+\dfrac{3}{4})+(sin^2B+\dfrac{3}{4})]+\dfrac{\sqrt{3}}{2}[(\dfrac{sin^2A}{3}+cos^2B)+(\dfrac{sin^2B}{3}+cos^2A)]=\dfrac{3\sqrt{3}}{2}$
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