Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$ 
How to prove that $\forall x \in \mathbb{R}$, $n \in \mathbb{N}$, we have
  \begin{align}
\sum_{k=1}^{n}\frac{|\sin{kx}|}{k}\ge |\sin{nx}| ?
\end{align}

I know that this partial sum will diverge for $x\not = m\pi$, but I don't know how to prove this inequality. I have tried Abel summation, but it doesn't work because I can't give a lower bound for $\sum |\sin{kx}|$. Thanks for your attention.
 A: Given $n\ge 1$ and $x\in \Bbb R$, denote 
$$f_n(x)=\sum_{k=1}^n\frac{|\sin kx|}{k} \quad\text{and}\quad  g_n(x)=f_n(x)-|\sin nx|.$$


Lemma: For every $n\ge 1$,
(i) $f_n$ is increasing on $[0,\frac{\pi}{n+1}]$;
   (ii) $g_n$ is increasing on $[0,\frac{\pi}n]$;
   (iii) $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.

Proof of Lemma: Note that when $x\in [0,\frac{\pi}n]$, 
$$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad\text{and}\quad g_n(x)=f_n(x)-\sin nx,$$
so 
$$f_n'(x)=\sum_{k=1}^n \cos kx \quad\text{and}\quad g_n'(x)=f_n'(x)-n\cos nx.$$
(i) Given $x\in[0,\frac{\pi}{n+1}]$, noting that $\cos\frac{kx}{2}\ge 0$ for $k=0,\pm1,\dots, \pm (n+1)$, we have
$$f_n'(x)=\sum_{k=1}^n \frac{\cos kx +\cos(n+1-k)x}{2}=\cos \frac{(n+1)x}{2} \cdot\sum_{k=1}^n \cos\frac{(n+1-2k)x}{2}\ge 0.$$
(ii) Since the cosine function is decreasing on $[0,\pi]$, when $x\in [0,\frac{\pi}{n}]$,  $\cos k x\ge \cos nx$ for $k=1,\dots,n$, so $g'(x)\ge 0$.
(iii) When $n=1$,  the statement is clearly true; when $n=2$, since $f_2$ is concave on $[\frac{\pi}{4},\frac{\pi}{2}]$, $f_2(\frac{\pi}{4})>1$ and $f_2(\frac{\pi}{2})=1$, the statement is also true. By induction, we may assume that $f_{n-1}\ge 1$ on $[\frac{\pi}{2(n-1)},\frac{\pi}{2}]$ for some $n \ge 3$, and the conclusion $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$ follows from the facts below. Firstly, $f_n\ge f_{n-1}$; secondly, $f_n$ is increasing on $[0,\frac{\pi}{n+1}]\supset [\frac{\pi}{2n},\frac{\pi}{2(n-1)}]$; thirdly, 
$$\sin \frac{\pi t}{2}\ge t,\ \forall t\in[0,1]\Longrightarrow f_n(\frac{\pi}{2n})=\sum_{k=1}^n\frac{\sin \frac{k\pi}{2n}}{k}\ge 1.\qquad \square$$

Now we can prove that $g_n\ge 0$ by using the lemma. Since $g_n(\pi\pm x)=g_n(x)$, we may focus on $x\in[0,\frac{\pi}{2}]$. Since $g_n(0)=0$, by (ii), we know that $g_n(x)\ge 0$ on $[0,\frac{\pi}{n}]$. Since $g_n\ge f_n -1$, by (iii) we know that $g_n\ge 0$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.
A: Let's assume $nx\le\pi$. A glance at the graph of $\sin x$ shows that for $0\le k\le n$ the line through the origin and the point $(kx,\sin kx)$ passes through or above the point $(nx,\sin nx)$, so $\sin kx\ge(k/n)\sin nx$. So $$\sum_{k=1}^n{\sin kx\over k}\ge\sum_{k=1}^n{\sin nx\over n}=\sin nx$$ 
I'm confident that the general case is just a matter of filling in a few details. 
A: You can get a lower bound as follows:
$$
|\sin(k\,x)|\ge|\sin(k\,x)|^2=\frac{1-\cos(2\,k\,x)}{2}.
$$
From this it is easy to get
$$
\sum_{k=1}^n\frac{|\sin(k\,x)|}{k}\ge\sum_{k=1}^n\frac{1}{2\,k}-\sum_{k=1}^n\frac{\cos(2\,k\,x)}{2\,k}\ge\frac12\log n-C(x),\quad x\ne0,\pi,
$$
for some $C(x)>0$. This will prove the inequality for large values of $n$.
