A simple inequality in calculus? I have to solve this inequality:
$$\left(\left[\dfrac{1}{s}\right] + 1 \right)  s < 1,$$ 
where   $ 0 < s < 1 $.
I guess that $s$ must be in this range: $\left(0,\dfrac{1}{2}\right]$.But I do not know if my guess is true. If so, how I can prove it?
Thank you.

The question states that $s$ is positive hence $\left[\frac{1}{s}\right]+1<\frac{1}{s}$ and so $\left[\frac{1}{s}\right]<\frac{1}{s}-1$. Assuming that the square brackets represent the 'integer part' function then, as $s>0$, we also know that $\frac{1}{s}>0$ and so $\left[\frac{1}{s}\right]=\frac{1}{s}-\alpha$ for some $\alpha\in[0,1)$. This then gives us $\frac{1}{s}-\alpha<\frac{1}{s}-1\Rightarrow 1<\alpha$ which is a contradiction, and so no $s$ in the range $(0,1)$ exists which satisfies the inequality.
If the square bracket notation does not mean 'integer part of' then please explain the notation better.
