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
 A: $\sin(a) + \sin(b) > \sin(a+b)$ if $(a,b)\in (0, \pi)\implies $ 
$\sin(a + b +c ) <= \sin(a) + \sin(b + c) < \sin(a) + \sin(b) + \sin(c)$
A: 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 $$\sin (a+b+c)<\sin a+\sin b+\sin c\Rightarrow \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$$
A: Let $\sin (a)=x,\sin (b)=y,\sin (c)=z $ thus continuing from your simplified version we have $$\frac {\sum ^{cyc} x\sqrt {(1-y^2)(1-z^2)}}{x+y+z}$$ then using Am-Gm for each sqyare root we have $$\frac {\sum^{cyc} x\sqrt{(1-y^2)(1-z^2)}}{x+y+z }\leq \frac {2 (x+y+z)-2 (xy+yz+xz)}{2 (x+y+z)} $$ thus its $1-\bf{something} $ we can easily see that the second bracket has maximum value $1$
but in the problem as $90^{0}$ is not in domain so its always less than $1$ 
Thus the original expression is less than $1$
