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If $a,b,c\in \left(0,\frac{\pi }{2}\right)\phantom{\rule{thickmathspace}{0ex}},$, Then prove that $\frac{\mathrm{sin}\left(a+b+c\right)}{\mathrm{sin}a+\mathrm{sin}b+\mathrm{sin}c}<1$
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Jayvion Tyler
$\mathrm{sin}\left(a\right)+\mathrm{sin}\left(b\right)>\mathrm{sin}\left(a+b\right)$ if $\left(a,b\right)\in \left(0,\pi \right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}$
$\mathrm{sin}\left(a+b+c\right)<=\mathrm{sin}\left(a\right)+\mathrm{sin}\left(b+c\right)<\mathrm{sin}\left(a\right)+\mathrm{sin}\left(b\right)+\mathrm{sin}\left(c\right)$

pipantasi4
Using
$\mathrm{sin}\left(a+b+c\right)-\mathrm{sin}a-\mathrm{sin}b-\mathrm{sin}c$
$=2\mathrm{cos}\left(\frac{2a+b+c}{2}\right)\mathrm{sin}\left(\frac{b+c}{2}\right)-2\mathrm{sin}\left(\frac{b+c}{2}\right)\mathrm{cos}\left(\frac{b-c}{2}\right)$
So
$=2\mathrm{sin}\left(\frac{b+c}{2}\right)\left[\mathrm{cos}\left(\frac{2a+b+c}{2}\right)-\mathrm{cos}\left(\frac{b-c}{2}\right)\right]$
$=-4\mathrm{sin}\left(\frac{a+b}{2}\right)\mathrm{sin}\left(\frac{b+c}{2}\right)\mathrm{sin}\left(\frac{a+c}{2}\right)<0,$
Bcz given $a,b,c\in \left(0,\frac{\pi }{2}\right)$. So we get $\frac{a+b}{2},\frac{b+c}{2}\phantom{\rule{thickmathspace}{0ex}},\frac{c+a}{2}\in \left(0,\frac{\pi }{2}\right)$
So we get
$\mathrm{sin}\left(a+b+c\right)<\mathrm{sin}a+\mathrm{sin}b+\mathrm{sin}c⇒\frac{\mathrm{sin}\left(a+b+c\right)}{\mathrm{sin}a+\mathrm{sin}b+\mathrm{sin}c}<1$