$$\left| \sin \left( \sum_{k=1}^n x_k \right) \right| \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$
I know it is true. Prove it...True.
(Trig. identities)
What has that got to do with the price of tea in China?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.
How much does tea cost in China? It is sufficient.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 $x_k$.
Since \(\displaystyle \frac{f(1+n) }{f(n)} >1\) then \(\displaystyle f(n)\uparrow\) and min{\(\displaystyle n\cdot \sin(\pi /2n )\)}\(\displaystyle =1\geq 1\).$ \displaystyle f(n)=n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N} $, remains to prove the inequality.
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