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Question Number 84849 by M±th+et£s last updated on 16/Mar/20

$$ABCisatriangle$$ $$provethat$$ $$sinA+sinB+sinC>sinAsinBsinC$$

Commented byajfour last updated on 16/Mar/20

$$0<\sin A,\sin B,\sin C⩽1$$ $$soitshouldbeobvious.$$

Commented byM±th+et£s last updated on 16/Mar/20

$$thatisrightbutineedaprove$$

Commented byMJS last updated on 17/Mar/20

$$\sin \alpha +\sin \beta +\sin \gamma$$ $$\sin \alpha +\sin \beta +\sin (\pi -(\alpha +\beta ))$$ $$\sin \alpha +\sin \beta +\sin (\alpha +\beta )$$ $$2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+\sin (\alpha +\beta )$$ $$2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$2\sin \frac{\alpha +\beta }{2}(\cos \frac{\alpha -\beta }{2}+\cos \frac{\alpha +\beta }{2})$$ $$2\sin \frac{\alpha +\beta }{2}\left(2\cos \frac{\frac{\alpha -\beta }{2}+\frac{\alpha +\beta }{2}}{2}\cos \frac{\frac{\alpha -\beta }{2}-\frac{\alpha +\beta }{2}}{2}\right)$$ $$4\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha }{2}\cos \frac{\beta }{2}$$ $$$$ $$\sin \alpha \sin \beta \sin \gamma$$ $$\sin \alpha \sin \beta \sin (\alpha +\beta )$$ $$\sin \left(2\frac{\alpha }{2}\right)\sin \left(2\frac{\beta }{2}\right)\sin \left(2\frac{\alpha +\beta }{2}\right)$$ $$2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}2\sin \frac{\beta }{2}\cos \frac{\beta }{2}2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$8\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$$$ $$4\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha }{2}\cos \frac{\beta }{2}>8\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha }{2}\cos \frac{\beta }{2}>2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$0<\alpha +\beta <\pi \Rightarrow \sin \frac{\alpha +\beta }{2}>0$$ $$\cos \frac{\alpha }{2}\cos \frac{\beta }{2}>2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$0<\alpha <\pi \Rightarrow 0<\frac{\alpha }{2}<\frac{\pi }{2}\Rightarrow \cos \frac{\alpha }{2}>0\mathrm{same}\mathrm{for}\beta$$ $$1>2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\alpha +\beta }{2}$$ $$1>2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2})$$ $$1>2(\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}-{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2})$$ $$1>2(\frac{1}{2}\sin \alpha \sin \beta -{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2})$$ $$1>\sin \alpha \sin \beta -\frac{1}{2}{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2}$$ $$0<\sin \alpha \sin \beta <1\wedge 0<\frac{1}{2}{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2}<1$$ $$\Rightarrow \mathrm{true}$$

Commented byM±th+et£s last updated on 17/Mar/20

$$godblessyousir$$

Commented byMJS last updated on 17/Mar/20

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$ $$\mathrm{maybe}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{shorter}\mathrm{way},\mathrm{I}\mathrm{just}\mathrm{started}$$ $$\mathrm{trying}\mathrm{and}\mathrm{this}\mathrm{is}\mathrm{where}\mathrm{it}\mathrm{lead}\mathrm{me}...$$

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