If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$ 
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$

$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$
$$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$
$$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$
Now how can i solve it after  that , Help required, Thanks

Using $$\sin (a+b+c)-\sin a-\sin b-\sin c $$
$$= 2\cos\left(\frac{2a+b+c}{2}\right)\sin \left(\frac{b+c}{2}\right)-2\sin \left(\frac{b+c}{2}\right)\cos \left(\frac{b-c}{2}\right)$$
So $$ = 2\sin \left(\frac{b+c}{2}\right)\left[\cos \left( \frac{2a+b+c}{2}\right)-\cos \left(\frac{b-c}{2}\right)\right]$$
$$ = -4\sin \left(\frac{a+b}{2}\right)\sin \left(\frac{b+c}{2}\right)\sin \left(\frac{a+c}{2}\right)<0,$$
Bcz given $\displaystyle a,b,c \in \left(0,\frac{\pi}{2}\right)$. So  we get $\displaystyle \frac{a+b}{2},\frac{b+c}{2}\;,\frac{c+a}{2}\in \left(0,\frac{\pi}{2}\right)$
So we get $$\sin (a+b+c)<\sin a+\sin b+\sin c\Rightarrow \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$$
