Why can no task have a utilization rate greater than one?
It appears that you have a periodic scheduling system on a single processor where task&nbsp;$i$ must run for $C_i$&nbsp;seconds out of every&nbsp;$T_i$.

If this is the case, then the requirement that $\tfrac{C_i}{T_i}\le1$ is just saying that $C_i\le T_i$, i.e., you're not saying something like "Task&nbsp;3 must run for 15&nbsp;seconds out of every 10&nbsp;seconds", which is physically impossible.

It looks like there's actually a mistake in the statement of the theorem. I think it should say that the utilization of task&nbsp;$i$ is given by $U_i=\tfrac{C_i}{T_i}$, then define the total utilization to be $U = \sum_{i=1}^n U_i$ and then require the stronger condition that this&nbsp;$U$ satisfies $U\leq 1$. That is, you don't just require each task individually to use at most 100% of your resources: you require the whole ensemble to use at most 100%. You can think of it this way: if I'm supposed to spend $p_1\%$ of my time on task&nbsp;1, ..., and $p_n\%$ of my time on task&nbsp;$n$, I'd better not be committing more than 100% of my time!

The physical significant of defining $\hat{T} = \mathrm{lcm}(T_1, \dots, T_n)$ is that it's the shortest period of time in which every task is supposed to run an integer number of times. It's not essential that you consider the shortest such period: the proof would work fine if you used any period in which every task needs to run an integer number of times. Perhaps it would be conceptually easier to set $\hat{T} = \prod_{i=1}^n T_i$ instead.