6

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

juantheron
  • 53,015

3 Answers3

4

$\sin(a) + \sin(b) > \sin(a+b)$ if $(a,b)\in (0, \pi)\implies $

$\sin(a + b +c ) <= \sin(a) + \sin(b + c) < \sin(a) + \sin(b) + \sin(c)$

kotomord
  • 1,814
  • Thanks kotomord, In $\displaystyle x\in \left(0,\frac{\pi}{2}\right),$ Here $\sin (a+b) = \sin a\cos b +\cos a \sin b>\sin a+\sin b,$ But i did not understand how first line is true for $x\in (0,\pi)$ also did not understand second line. – juantheron Nov 01 '16 at 13:07
  • 1
    Theres a mistake in the first line of your soln. The inequality should be reversed, ie , sin(a) + sin(b) > sin(a+b). – SirXYZ Nov 01 '16 at 13:18
  • $\sin(\pi/3)+\sin(\pi/3)=\sqrt{3}>\sin(\pi/3+\sin\pi/3)=\sqrt{3}/2$. – egreg Nov 01 '16 at 13:22
  • Thanks egreg and SirJPM got it, But still did not understand how $\sin a+\sin b>\sin(a+b)\forall a,b\in (0,\pi)$ – juantheron Nov 01 '16 at 13:39
  • How does the second line follow from the first (corrected) one? This is completely unjustified. – egreg Nov 01 '16 at 13:41
  • 1
    sin(a)[1-cos(b)] + sin(b)[1-cos(a)] >=0 ..... for all (a,b) belonging to (0°,180°) .....since sin(a) and sin(b) will both be positive in the given interval and the min. value of the given expression will be 0 (when both sin(a) and sin(b) are equal to 0). Now from this inequality....you can easily derive the inequality sin(a)+sin(b)>sin(a+b). – SirXYZ Nov 01 '16 at 13:56
4

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 $$\sin (a+b+c)<\sin a+\sin b+\sin c\Rightarrow \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$$

juantheron
  • 53,015
2

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 $1$

juantheron
  • 53,015
Archis Welankar
  • 15,910
  • To Archis Welankar I did not understand how can you apply $\bf{A.M\geq G,M}$ Inequality in $\displaystyle \frac {\sum ^{cyc} x\sqrt {(1-y^2)(1-z^2)}}{x+y+z}$ – juantheron Nov 01 '16 at 13:48
  • No basically have just applied like$\sqrt {(1-y^2)(1-z^2)} \leq $\frac {2-(y+z)}{2} $ similarly for other two terms and then continued – Archis Welankar Nov 01 '16 at 14:01
  • Nice solution Archis Welankar, may be you mean $\displaystyle \sum ^{\cyc}x\sqrt{(1-y^2)(1-z^2)}<\sum ^{\cyc}x,$ bcz $\displaystyle x,y,z \in \left(0,1\right)$ – juantheron Nov 10 '16 at 13:59
  • Thanks and Yes I mean the same thing – Archis Welankar Nov 10 '16 at 16:06