Is there any way to prove this? When I check intuitively by taking $A=0$ and $B=C=\frac{\pi}{2}$, the value of the expression becomes $2$ and as I changed the angles the value kept increasing in $0$ to $\frac{\pi}{2}$. I tried using Jensen's inequality but we get $\sin A + \sin B + \sin C <\frac{3\sqrt{3}}{2}$, which is of no use.
Asked
Active
Viewed 606 times
1
-
Try using $$\sin A + \sin B + \sin C = 4\cos\left(\frac A 2 \right)\cos\left(\frac B 2 \right)\cos\left(\frac C 2 \right)$$ and see what happens. – stoic-santiago Apr 23 '21 at 04:45
-
Thanks for the reply, I tried using that but it goes into a loop. I was not able to go forward. – Purnima Kompella Apr 23 '21 at 05:01
-
Check what happens if $A$ and $B$ are close to $0$ and $C$ is close to $\pi$. – trancelocation Apr 23 '21 at 05:23
-
Try Karamata Inequality – Hari Shankar Apr 23 '21 at 07:18
-
Both duplicate targets found with Approach0: https://approach0.xyz/search/?q=%24%5Csin(a)%2B%5Csin(b)%2B%5Csin(c)%5Cge%202%24 – Martin R Apr 23 '21 at 08:05
2 Answers
0
$$\sin A + \sin B + \sin C=\sin A +\sin B + \sin(\pi-A-B)$$ $$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\sin(A+B)$$ $$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+2\sin(\frac{A+B}{2})\cos(\frac{A+B}{2})$$ $$=2\sin(\frac{A+B}{2})\Bigl(\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\Bigr)$$ $$=4\sin(\frac{A+B}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$ $$=4\sin(\frac{\pi-C}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$ $$=4\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}$$ $$=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$ $$0<A<B<\frac{\pi}{2}$$ $$0<\frac{A}{2}<\frac{B}{2}<\frac{\pi}{4}$$ $$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{C}{2}$$
when $A=0, B=0$. During this situation, $C=\frac{\pi}{2}$
So, $$\sin A + \sin B + \sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{\frac{\pi}{2}}{2}=2\sqrt{2}>2$$
Smriti Sivakumar
- 1,009
-
1Use
\left(and\right)for adjusting the size of brackets. For example, $$(\frac69)\text{ vs }\left(\frac69\right)$$ – ultralegend5385 Apr 23 '21 at 05:18 -
-
One more thing: when nesting brackets, you can use
\big,\Bigand\Biggalso. For example, $$\Bigg(\Big(\big((x)\big)\Big)\Bigg)$$ – ultralegend5385 Apr 23 '21 at 09:50
0
I try to use comment but failed. so I just comments here. For a acute triangle, we must have all angles less than $\frac{\pi}{2}$, so $4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} $ can be direct to get the result.
chenbai
- 7,681