How to deduce the following trig relation?

Question

How can I deduce:

$$\sqrt{|x|}\sin(\frac{1}{x}) \le \sqrt{|x|}$$??

I know of the relation.

$$\sin(u) \le u$$ $$u = \frac{1}{x}$$ $$\sin(1/x) \le \frac{1}{x}$$

But nothing related to $\sqrt{x}$

Thanks!

Answer 1

Since we have for $x\not=0\in\mathbb R$ $$\sin\left(\frac 1x\right)\le 1,$$ multiplying the both sides by $\sqrt{|x|}\gt 0$ gives us $$\sqrt{|x|}\sin\left(\frac 1x\right)\le \sqrt{|x|}.$$

Answer 2

Hint:
You have that $$\sin u\leqslant 1\quad\text{for all $u\in\Bbb R$},$$ and see what happens if you multiply both sides with $\sqrt{|x|}$.

Answer 3

It follows from the fact that $\sin(y)\leq 1, \forall y \in \mathbb{R}$. Hence you have

$$\sqrt{|x|}\sin(\frac{1}{x}) \le \sqrt{|x|} \times 1 = \sqrt{|x|}$$

As desired.

Answer 4

For any $\;x\neq 0\;$ (obviously), we have

$$\sqrt{|x|}\sin\frac1x\le\sqrt{|x|}\iff \sin\frac1x\le \frac{\sqrt{|x|}}{\sqrt{|x|}}= 1$$

which is trivial.