As my further preparation to Putnam competition, I came across such inequality to prove:
$$\forall n \in \mathbb{N}: \int_{0}^{\pi} \left|\frac{\sin(nx)}{x}\right|dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}$$
The problem is that I got stuck in spite of spending by far 6 days on this problem!
Because I got really stuck, I am very determined to see how to prove such inequality. Help very, very appreciated!
We may start from:
$$ A_k = \int_{0}^{\pi}\frac{\sin x}{x+k\pi}\,dx = \int_{0}^{\pi/2}\sin(x)\left(\frac{1}{x+k\pi}+\frac{1}{\pi-x+k\pi}\right)\,dx$$ and notice that $f_k(x)=\frac{1}{x+k\pi}+\frac{1}{\pi-x+k\pi}$ is decreasing on $\left[0,\frac{\pi}{2}\right]$, hence: $$ A_k \geq \int_{0}^{\pi/2} \sin(x)\,f_k\left(\frac{\pi}{2}\right)\,dx = \frac{4}{(2k+1)\pi}\geq \frac{4}{(2k+2)\pi}=\frac{2}{\pi}\cdot\frac{1}{k+1}.$$ Then we may notice that: $$ \int_{0}^{\pi}\left|\frac{\sin(nx)}{x}\right|\,dx = \sum_{k=0}^{n-1}A_k \geq \color{red}{\frac{2}{\pi} H_n} $$ as wanted. My approach indeed proves a slightly stronger inequality, i.e.:
$$ \int_{0}^{\pi}\left|\frac{\sin(nx)}{x}\right|\,dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k\color{red}{-\frac{1}{2}}}.$$
We have $$\begin{align} \int_{0}^{\pi}\left|\frac{\sin\left(nx\right)}{x}\right|dx\stackrel{nx=v}{=} & \int_{0}^{n\pi}\left|\frac{\sin\left(v\right)}{v}\right|dv \\ = & \sum_{k=0}^{n-1}\int_{k\pi}^{\left(k+1\right)\pi}\left|\frac{\sin\left(v\right)}{v}\right|dv \\ = & \sum_{k=0}^{n-1}\left(-1\right)^{k}\int_{k\pi}^{\left(k+1\right)\pi}\frac{\sin\left(v\right)}{v}dv \\ \geq & \frac{1}{\pi}\sum_{k=0}^{n-1}\frac{\left(-1\right)^{k}}{k+1}\int_{k\pi}^{\left(k+1\right)\pi}\sin\left(v\right)dv \\ = & \frac{1}{\pi}\sum_{k=0}^{n-1}\frac{\left(-1\right)^{k}\left(-\cos\left(\left(k+1\right)\pi\right)+\cos\left(k\pi\right)\right)}{k+1} \\ = & \frac{2}{\pi}\sum_{k=0}^{n-1}\frac{1}{k+1}. \end{align}$$ Note that the third line is positive since $\sin\left(x\right)<0 $ if $x\in\left(k\pi,\left(k+1\right)\pi\right) $ and $k$ odd. And $$-\cos\left(\left(k+1\right)\pi\right)+\cos\left(k\pi\right)=2\left(-1\right)^{k}. $$
Hint:
$$\int_{(k-1)\pi}^{k\pi} \left| \frac{\sin x}{x} \right|\,dx \geq \int_{(k-1)\pi}^{k \pi} \frac{|\sin x|}{k \pi}$$
