$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$ 
Question: Show that $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right).$$

My approach: We know that $$1+\frac{1}{2}+\cdots+\frac{1}{n}=\int_0^1\left(1+x+x^2+\cdots+x^{n-1}\right)dx=\int_0^1\frac{x^n-1}{x-1}dx.$$
Therefore we have $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right) \\ \Leftrightarrow \int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx.$$
Now substituting $x=\pi t$, we have $$\frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx=\frac{2}{\pi^{n+1}}\int_0^\pi \frac{t^n-\pi^n}{t-\pi}dt.$$
Thus we have $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\int_0^1\frac{x^n-1}{x-1}dx \\ \Leftrightarrow \int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi^{n+1}}\int_0^\pi \frac{x^n-\pi^n}{x-\pi}dx\hspace{0.5 cm}...(1)$$ 
Therefore, if we show that $(1)$ is true, then we are done. Can someone provide me a hint?

Let $t_k=\frac{k\pi}{n}$ for $k=0,1,2,\ldots,n$.  We have
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx=\int_0^\pi\frac{\sin u}{(k-1)\pi+u}du,$$
where $u=nx-(k-1)\pi$.  For $0\le u \le \pi$, $(k-1)\pi \le (k-1)\pi+u\le k\pi$, so
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx\ge \int_0^\pi\frac{\sin u}{k\pi}du=\frac{1}{k\pi}\big(-\cos u\big)\Big|_0^\pi=\frac{2}{k\pi}.$$
Hence
$$\int_0^\pi\left|\frac{\sin nx}{x}\right|dx=\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx\ge \sum_{k=1}^n\frac{2}{k\pi},$$
which is what we want.  In fact, we have
$$\frac{2}{\pi}\sum_{k=1}^n\frac1k\le \int_0^\pi\left|\frac{\sin nx}{x}\right|dx \le \operatorname{Si}(\pi)+\frac{2}{\pi}\sum_{k=1}^{n-1}\frac1k,$$
where $\operatorname{Si}$ is the sine integral ($\operatorname{Si}(\pi)\approx1.8519$).
