How to prove floor function inequality $\sum\limits_{k=1}^{n}\frac{\{kx\}}{\lfloor kx\rfloor }<\sum\limits_{k=1}^{n}\frac{1}{2k-1}$ for $x>1$ 
Let $x>1$ be a real number. Show that for any positive $n$
$$\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }<\sum_{k=1}^{n}\dfrac{1}{2k-1}\tag{1}$$
  where $\{x\}=x-\lfloor x\rfloor$

My attempt: I try use induction prove this inequality.
It is clear for $n=1$, because $\{x\}<1\le \lfloor x\rfloor$.
Now if assume that $n$ holds, in other words: $$\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }<\sum_{k=1}^{n}\dfrac{1}{2k-1}$$
Consider the case $n+1$. We have
$$\sum_{k=1}^{n+1}\dfrac{\{kx\}}{\lfloor kx\rfloor }=\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }+\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}<\sum_{k=1}^{n}\dfrac{1}{2k-1}+\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}$$
It suffices to prove that
$$\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}<\dfrac{1}{2n+1}\tag{2}$$
But David gives an example showing $(2)$ is wrong, so how to prove $(1)$?
 A: Some thoughts:
Looking at several plots indicates that $$f_n(x):=\sum_{k=1}^n{\{kx\}\over\lfloor kx\rfloor}$$
is largest immediately to the left of $x=2$. Now for $x=2-\epsilon$ with $0<\epsilon\ll1$ one has
$$\lfloor kx\rfloor=2k-1,\quad\{kx\}=1-2k\epsilon$$
and therefore
$$f_n(x)=\sum_{k=1}^n{1-2k\epsilon\over2k-1}<\sum_{k=1}^n{1\over2k-1}\ .$$
Maybe you want to take a look at the following graph of $f_{250}$:

A: Equation (1) is not true in general, in fact, for every $n$ one can find an $x$ for which it is false.  Specifically, given $n\ge1$, choose
$$x=\frac{n+\frac74}{n+1}>1\ .$$
Then $2n>1$, so $4n+4=4n+3+1<6n+3$, so
$$\frac{\{(n+1)x\}}{\lfloor(n+1)x\rfloor}
  =\frac{\frac34}{n+1}=\frac3{4n+4}>\frac3{6n+3}=\frac1{2n+1}\ .$$
A: I tried all day and couldn't prove it but I made a little progress:
Let's define $\{x\}'$ to be 1 if $x$ is an integer and $\{x\}$ otherwise, and note that the LHS of the original inequality satisfies
$$\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor} \leq \sum_{k=1}^{n}\dfrac{\{kx\}'}{\lceil kx\rceil-1}\tag{1}$$
If $a=\lceil nx\rceil$ then $\lceil kx\rceil=\lceil k\frac an\rceil$ for $k=1,2,... n$ (can be proved by contradiction), and the modified fractional part is non-decreasing, so it suffices to prove that 
$$\sum_{k=1}^{n}\dfrac{\{k\frac an\}'}{\lceil k\frac an\rceil-1}<\sum_{k=1}^{n}\dfrac{1}{2k-1}$$
for integers $a\in (n,2n)$ (since we can assume $1<x<2$).
The RHS of (1) can be rewritten as
$$\sum_{k=1}^{n}\dfrac{\{kx\}'}{kx-\{kx\}'}=\sum_{k=1}^{n}\dfrac{1}{kx/\{kx\}'-1}$$
since $\lceil kx\rceil=kx+(1-\{kx\}')$.
Letting $x=\frac an$, if $\gcd(a,n)=1$ then $\{\{kx\}':k=1,2,...n\}=\{\frac 1n,\frac 2n,...\frac nn\}$. For $k\in[1,n-1]$, let unique $t\in[1,n]$ such that $t\equiv ak\pmod{n}$. Then $\{kx\}'=\frac tn$ and $k=[a^{-1}t]_n$ so we can write our summation with index $t$:
$$\frac1{a-1}+\sum_{k=1}^{n-1}\dfrac{1}{kx/\{kx\}'-1}=\frac1{a-1}+\sum_{t=1}^{n-1}\dfrac{1}{k\frac an/\frac tn-1}$$
$$=\frac1{a-1}+\sum_{t=1}^{n-1}\dfrac{1}{k\frac at-1}$$
Now since $ka\equiv t\pmod n$ we have $ka=u_tn+t$ for some $u_t\in[1,a]$, so this then becomes
$$=\frac1{a-1}+\sum_{t=1}^{n-1}\dfrac{1}{\frac {u_tn+t}t-1}$$
$$=\frac1{a-1}+\frac 1n\sum_{t=1}^{n-1}\dfrac{t}{u_t}$$
I stopped at this point but I'll try to see if I can turn it into a proof tomorrow. Comment if you have any ideas!

