Inequality

tahirimanov19

$$\left| \sin \left( \sum_{k=1}^n x_k \right) \right| \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$

DarnItJimImAnEngineer

True.

(Trig. identities)

DarnItJimImAnEngineer

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.

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idontknow

I will post the solution if I prove it.
Can you give at least a hint?

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idontknow

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.

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DarnItJimImAnEngineer

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 $x_k$.

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tahirimanov19

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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idontknow

$\displaystyle n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N}$, remains to prove the inequality.

idontknow

$\displaystyle f(n)=n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N}$, remains to prove the inequality.
Since $$\displaystyle \frac{f(1+n) }{f(n)} >1$$ then $$\displaystyle f(n)\uparrow$$ and min{$$\displaystyle n\cdot \sin(\pi /2n )$$}$$\displaystyle =1\geq 1$$.