$\int_0^1 |\sin n x| dx \ge C >0$? I want to prove that 
$$
\int_0^1 |\sin n x| dx \ge C >0
$$
for any $n\in \mathbb N$ with $C$ not depending on $n$.
This seems so difficult for me to prove.
 A: $\newcommand{\abs}[1]{\left\lvert{#1}\right\rvert}\newcommand{\integ}[1]{\left\lfloor {#1}\right\rfloor}$
We can safely assume $n\ge2$.
\begin{align}
\int_0^1\abs{\sin(nx)}\,dx&=\frac1n\int_0^n\abs{\sin x}\,dx=\frac1n\left(\int_0^{\frac\pi2\integ{\frac{2n}\pi}}\abs{\sin x}\,dx+\int_{\frac\pi2\integ{\frac{2n}\pi}}^n\abs{\sin x}\,dx\right)\ge\\&\ge\frac1n\int_0^{\frac\pi2\integ{\frac{2n}\pi}}\abs{\sin x}\,dx=\left(\text{because of }\pi\text{-periodicity and symmetries}\right)\\&=\frac{1}{n}\integ{\frac{2n}\pi}\int_0^{\pi/2}\abs{\sin x}\,dx
=\frac1n\integ{\frac{2n}\pi}=\frac{1}{n}\integ{\frac{2/\pi}{1/n}}\ge \frac 2\pi-\frac1n\end{align}
Since for positive real numbers $b\cdot\integ{\frac{a}{b}}\ge a-b$.
For $n\ge2$, that quantity is $\ge \dfrac2\pi-\dfrac12=\dfrac{4-\pi}{2\pi}>0$
A: Since $0 \leq |\sin (nx)| \leq 1$, we have
$$
0 \leq \sin^{2}(nx) = |\sin(nx)|^{2} \leq |\sin(nx)| \leq 1
$$
for all $n$, and all real $x$. Consequently,
\begin{align*}
\int_{0}^{1} |\sin(nx)|\, dx
  &\geq \int_{0}^{1} \sin^{2}(nx)\, dx
   = \frac{1}{n} \int_{0}^{n} \sin^{2} x\, dx
   = \frac{1}{2n} \int_{0}^{n} [1 - \cos(2x)]\, dx \\
  &= \frac{1}{2n} \left[x - \frac{\sin(2x)}{2}\right]\bigg|_{0}^{n}
   = \frac{1}{2n} \left[n - \frac{\sin(2n)}{2}\right]
   = \frac{1}{2} - \frac{\sin(2n)}{4n} \\
  &\geq \frac{1}{2} - \frac{1}{4n}
   \geq \frac{1}{4}.
\end{align*}
A: $$\int_0^1|\sin(nx)|dx = \frac{1}{n}\int_0^{n}|\sin y|dy = \frac{1}{n}\sum_{i = 1}^n\int_{i-1}^i|\sin y|dy \ge \frac{1}{n}n\min_{i \in \mathbb{N}}{\int_{i-1}^i|\sin y|dy}.$$
We can now show that 

Claim: $$\min_{i \in \mathbb{N}}{\int_{i-1}^i|\sin y|dy} \ge \int_{-\frac 12}^{\frac 12}|\sin y|dy = C.$$

Indeed, $$\frac{d}{dx}\int_x^{x + 1}|\sin y|dy = |\sin(x + 1)| - |\sin x|$$ and the RHS is $0$ if and only if $x + \frac 12 \in \{k\frac{\pi}{2} : k \in \mathbb{Z}\}$. Clearly, $k$ even corresponds to a minimum while $k$ odd corresponds to a maximum so that taking $k = 0$ and $x = -\frac 12$ proves the claim.
To compare this with the other answers, here are the values of $C$ that we have found so far:


*

*G. Sassatelli: 0.136619

*Andrew D. Hwang: 0.25

*Giovanni: 0.24483


In conclusion, Andrew's estimate is the sharpest so far :)
