Inequality

Mar 2015
182
68
Universe 2.71828i3.14159
$$\left| \sin \left( \sum_{k=1}^n x_k \right) \right| \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$
 
Jun 2019
493
262
USA
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:
Dec 2015
1,084
169
Earth
I will post the solution if I prove it.
Can you give at least a hint?
 
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Dec 2015
1,084
169
Earth
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 a moderator:
Jun 2019
493
262
USA
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.
What has that got to do with the price of tea in China?
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 a moderator:
Mar 2015
182
68
Universe 2.71828i3.14159
What has that got to do with the price of tea in China?
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 $x_k$.
How much does tea cost in China? It is sufficient.
 
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Dec 2015
1,084
169
Earth
$ \displaystyle n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N} $, remains to prove the inequality.
 
Dec 2015
1,084
169
Earth
$ \displaystyle f(n)=n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N} $, remains to prove the inequality.
Since \(\displaystyle \frac{f(1+n) }{f(n)} >1\) then \(\displaystyle f(n)\uparrow\) and min{\(\displaystyle n\cdot \sin(\pi /2n )\)}\(\displaystyle =1\geq 1\).