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October 16th, 2019, 01:35 PM   #1
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Inequality

$$\left| \sin \left( \sum_{k=1}^n x_k \right) \right| \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$
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October 16th, 2019, 02:26 PM   #2
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True.

(Trig. identities)
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October 16th, 2019, 03:32 PM   #3
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True.

(Trig. identities)
I know it is true. Prove it...
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October 16th, 2019, 03:45 PM   #4
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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.
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October 17th, 2019, 01:16 AM   #5
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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.
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October 17th, 2019, 04:32 AM   #6
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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.
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October 17th, 2019, 11:21 AM   #7
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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 skipjack; October 18th, 2019 at 09:58 PM.
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October 17th, 2019, 11:50 AM   #8
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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.

Last edited by skipjack; October 18th, 2019 at 09:56 PM.
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October 18th, 2019, 05:08 AM   #9
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$ \displaystyle n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N} $, remains to prove the inequality.
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October 19th, 2019, 02:56 AM   #10
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$ \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{$\displaystyle n\cdot \sin(\pi /2n )$}$\displaystyle =1\geq 1$.
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