Is true that $\sum_{k=1}^n\frac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$? Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$?
My work:
Let $x\in\mathbb{R}$ be given. For every $k=1,\dots,n$, define $i_k\in\{0,1,\dots,k-1\}$ such that 
$$
[x]+\frac{i_k}{k}\leq x<[x]+\frac{i_k+1}{k}\hspace{20mm} (*).
$$
On one hand, we have that $[kx]=k[x]+i_k$ for every $k=1,\dots,n$.
So  $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}=n[x]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}=[nx]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}-i_n$
Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}\leq i_n\,.$$
On the other hand, from $(*)$ is direct that $i_k=[k(x-[x])]$.
Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{[k(x-[x])]}{k}\leq [n(x-[x])]\,.$$
In other words, we can assume that $x\in[0,1)$. 
Now the questions is:
Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in[0,1)$ and for every $n\in\mathbb{Z}^+$?
Any simpler approach? Any thoughts? 
 A: Note that the function
$$f:\mathbb{R}\to\mathbb{R},f(x)=[nx]-\sum_{k=1}^n\frac{[kx]}{k}$$
is right-continuous and $1$-periodic function, so it is enough to prove that $f(x)\ge0$ for $x\in[0,1)$. Note also that the set of points of discontinuity of $f$ on $(0,1)$ is contained in
$$\mathcal{D}=\left\{\frac{p}{q}: 1\le p< q\le n,\gcd(p,q)=1\right\}$$ and that $f$ is constant on any interval contained in $[0,1)\setminus\mathcal{D}$. Therefore, in order to prove that $f(x)\ge0$ on $(0,1)$ it is enough to prove that
$f(r)\ge0$ for $r\in\mathcal{D}$.
Now, consider $r=p/q\in\mathcal{D}$ with $1\le p<q\le n$ and $\gcd(p,q)=1$.
For $k\in\{1,\ldots,q-1\}$ we define
$\phi(k)=pk-q[kr]$. Since $[kr]\le kr<[kr]+1$ we see that $0\le\phi(k)<q$, moreover, if $\phi(k)=0$ for some $k$ we conclude that $q$ must divide $k$, beause
$\gcd(p,q)=1$, which is absurd. Thus $\phi$ takes its values in $\{1,\ldots,q-1\}$. Further, $\phi$ is one to one, because if $\phi(k)=\phi(\ell)$ for some
$1\le k\le\ell<q$ we conclude that $q|(p(\ell-k))$ and again, because $\gcd(p,q)=1$ we see that $q|(\ell-k)$ and this implies that $\ell=k$ because
$0\le \ell-k<q$. Thus, $\phi$ is a permutation on the set $\{1,\ldots,q-1\}$.
Using the AM-GM inequality we have
$$\frac{1}{q-1}\sum_{k=1}^{q-1}\frac{\phi(k)}{k}\ge\left(\prod_{k=1}^{q-1}\frac{\phi(k)}{k}\right)^{1/(q-1)}=1$$
because $\phi(1)\phi(2)\cdot\phi(q-1)=1\cdot 2\cdots(q-1)$.
Thus
$$\sum_{k=1}^{n}\frac{kp-q[kr]}{k}\ge\sum_{k=1}^{q-1}\frac{kp-q[kr]}{k}\ge q-1
\ge pn-q[nr]$$
Rearranging we get $f(r)\ge0$ which is the desired conclusion. The proof is complete.
A: $$[nx] \geqslant \sum\limits_{k=1}^n \frac{[kx]}{k} = \sum\limits_{k=1}^n \frac{kx-\{kx\}}{k} = nx - \sum\limits_{k=1}^n \frac{\{kx\}}{k}\Leftrightarrow$$
$$f(x):=\sum\limits_{k=1}^n \frac{\{kx\}}{k} \geqslant \{nx\}=:g(x)$$
$f(0)=\{0\}=0$.
$f,g$ at points of differentiability are growing at a rate of $n$. Hence if the interval $(a,b)$ functions $f,g$ differentiable and $f(a)\geqslant g(a)$, even on this entire interval $f(x)\geqslant g(x)$. It remains to prove the inequality gap in the points. Break points - all rational numbers with denominators $\leqslant n$: $x=\frac{a}{m}$, $a,m$ are relatively prime.
$$f(\frac{a}{m}) \geqslant \sum\limits_{k=1}^{m-1} \frac{\{ka/m\}}{k}$$
Numerator - kind of shot $\frac{1}{m},...,\frac{m-1}{m}$, denominators -$1,...,m$. The lowest value of the summation for all permutations of the numerator is obtained when the numerator is proportional to the denominator. So $f(\frac{a}{m})\geqslant \sum\limits_{k=1}^{m-1} \frac{k}{km}=1-\frac{1}{m}\geqslant\{n\frac{a}{m}\}$
