Is $\frac{\lfloor{x}\rfloor+1}{2} \le \lfloor\frac{x}{2}\rfloor + 1$ The answer seems to be yes.
Here's my reasoning:
For $x < 1$, $\frac{1}{2} < 1$
For $1 < x < 2$, $1 = 1$
For $2 \mid x$, $\frac{x}{2} + \frac{1}{2} < \frac{x}{2} + 1$
For $2 \mid x-1$, $\frac{x+1}{2} = \frac{x-1}{2} + 1$
For all other values, the value is equal to one of the above.
My reasoning is not very straight forward.  What would be a standard way to demonstrate this? 
Thanks,
-Larry

Set $n = \lfloor \frac{x}{2} \rfloor$. Then $x< 2n + 2$ and thus $\lfloor x \rfloor < 2n+2$ which implies $\lfloor x \rfloor \leq 2n+1$ as $\lfloor x \rfloor$ is an integer. This gives 
$$\frac{\lfloor x \rfloor + 1}{2} \leq \frac{2n+2}{2} = n+1 = \lfloor \frac{x}{2} \rfloor + 1$$
