TITLE: Showing $\{x\} + \{\frac{1}{x}\} \lt 1.5$ and other problems.
QUESTION [5 upvotes]: For any real number $x$, let $[x]$ be the greatest integer not exceeding $x$. We also define $\{x\}=x-[x]$. We now define the function:
$f(x)=\{x\}+\{\frac{1}{x}\}$.
(a) Prove that $f(x)<1.5$ for $x>0$, and $f(x)<2$ for $x<0$.
(b) Prove that there exists a real number $\alpha $ such that $f(\alpha)>\frac{ 4001}{2001}$.
Going through cases I am convinced of (a). But how do you prove it conclusively?
(b) presented more challenges, but do we show existence without finding such {a} value{s}? Or do we need to find  specific value of $\alpha$?

REPLY [4 votes]: If $x = n+f$, where $n \ge 1$ (wlog) and $0 \lt f \lt 1$, then $\{\frac{1}{x}\} = \frac{1}{x}$ then
$$\{x\} + \{\frac{1}{x}\} = f + \frac{1}{n+f}$$
which is an increasing (and continuous) function of $f$ in $(0, 1])$, for $n \ge 1$. 
This you can show by considering $f(t) = n+t + \frac{1}{n+t}$, which is just $t + \frac{1}{t}$ in disguise (and we have $t \ge 1$ for our purposes).
The maximum occurs when you choose $n=f=1$, and since $f$ is never really $1$, the inequality is strict.
A similar argument can probably be put forward for the case $x \lt 0$, and continuity arguments will show existence without having to construct one.