October 16th, 2019, 01:35 PM  #1 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 64 Math Focus: Area of Circle  Inequality
$$\left \sin \left( \sum_{k=1}^n x_k \right) \right \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$

October 16th, 2019, 02:26 PM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 
True. (Trig. identities) 
October 16th, 2019, 03:32 PM  #3 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 64 Math Focus: Area of Circle  
October 16th, 2019, 03:45 PM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 
sin(a+b) = sin(a)cos(b)+cos(a)sin(b) The cosines will have a magnitude less than or equal to 1. Whether they add or subtract, the sum will have a magnitude less than the sum of the sines. Last edited by DarnItJimImAnEngineer; October 16th, 2019 at 04:34 PM. 
October 17th, 2019, 01:16 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math 
I will post the solution if I prove it. Can you give at least a hint? Last edited by skipjack; October 18th, 2019 at 10:01 PM. 
October 17th, 2019, 04:32 AM  #6 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math 
The bound of LHS occurs for $\displaystyle x_k = \pi /2n$. $\displaystyle 1\leq n\sqrt[n]{\sin(x_1 ) \sin(x_2) \cdot ...\cdot \sin(x_n )}=n\sqrt[n]{\sin^{n}(\pi /2n ) }=n\sin(\pi /2n ).$ $\displaystyle n\sin(\pi /2n ) \geq 1 $ proves the inequality. Last edited by skipjack; October 18th, 2019 at 09:59 PM. 
October 17th, 2019, 11:21 AM  #7  
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202  Quote:
Also, set n=1, and it is clearly false. I just demonstrated that $\sin(x_1 + x_2) = c_1 \sin(x_1) + c_2 \sin(x_2)$, where $c_1, c_2 \in [1, 1]$. Is that not sufficient proof that $\sin(x_1 + x_2) \leq \sin(x_1) + \sin(x_2)$ if the sines are positive? Then you can extend this to $\sin((x_1 + x_2) + x_3) \leq (\sin(x_1) + \sin(x_2)) + \sin(x_3)$, and keep adding terms up to $x_k$. Last edited by skipjack; October 18th, 2019 at 09:58 PM.  
October 17th, 2019, 11:50 AM  #8  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 64 Math Focus: Area of Circle  Quote:
Last edited by skipjack; October 18th, 2019 at 09:56 PM.  
October 18th, 2019, 05:08 AM  #9 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math 
$ \displaystyle n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N} $, remains to prove the inequality.

October 19th, 2019, 02:56 AM  #10 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math  

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