Does $\int_{0}^{\pi/2n}\left|\frac{\sin(2n+1)t}{\sin t}\right|\text{d}t\leq \pi$ hold for $n \geq 2$? Today I am trying to prove an integral inequality:
$$\frac{1}{\pi}\int_{0}^{\pi/2}\left|\frac{\sin(2n+1)t}{\sin t}\right|\text{d}t<\frac{2+\ln n }{2}$$
where $n\geq 2$ and $n \in \Bbb{N}$.

First, use Jordan's inequality: $\sin t \geq \frac{2}{\pi}t, t\in [0,\frac{\pi}{2}]$, I divide $[0,\pi/2]$ into $[0,\pi/2n]$ and $(\pi/2n,\pi/2]$,and estimate respectively. First, $$\int_{\frac{\pi}{2n}}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)t}{\sin t}\right|\text{d}t\leq \frac{\pi}{2}\int_{\frac{\pi}{2n}}^{\frac{\pi}{2}}\frac{1}{t}\text{d}t=\frac{\pi}{2}\ln n$$

The rest is to prove $\int_{0}^{\pi/2n}\left|\frac{\sin(2n+1)t}{\sin t}\right|\text{d}t\leq \pi$. then I can complete the proof. But I can't work it out. Does this inequality hold:
$$\int_{0}^{\pi/2n}\left|\frac{\sin(2n+1)t}{\sin t}\right|\text{d}t\leq \pi $$? thanks very much.
 A: Let's analyze the situation. You divide the interval into short  $[0,a]$ and long  $[a,\pi/2]$. The integral over the long interval is bounded using the estimate $|\sin t|\ge \frac{2}{\pi}t$, which yields 
$$
\int_{a}^{\pi/2}\left|\frac{\sin(2n+1)t}{\sin t}\right|\,dt\le \int_a^{\pi/2} \frac{\pi }{2 t} \,dt = \frac{\pi}{2}\ln \frac{\pi/2}{a}
\tag{1} $$
On the short interval, use the inequality $|\sin m t|\le m|\sin t|$, which follows by induction from $|\sin (m+1)t| = |\sin t\cos mt+\sin mt \cos t|\le |\sin t| + |\sin mt|$.
$$
\int_{0}^{a}\left|\frac{\sin(2n+1)t}{\sin t}\right|\,dt\le \int_0^a (2n+1)\,dt = (2n+1)a \tag{2}
$$
It remains to choose $a$ so that the sum of right-hand sides of (1) and (2) is minimal. A short computation with the derivative points toward $a=\frac{\pi/2}{2n+1}$. This yields
$$
\int_{0}^{\pi/2}\left|\frac{\sin(2n+1)t}{\sin t}\right|\,dt 
\le  \frac{\pi}{2} \ln (2n+1) + \frac{\pi}{2}
$$
hence 
$$
\frac{1}{\pi}\int_{0}^{\pi/2}\left|\frac{\sin(2n+1)t}{\sin t}\right|\,dt 
\le  \frac{1+\ln(2n+1) }{2}  
\tag{3}$$
Inequality (3) implies the estimate you wanted, because $$\ln(2n+1)<\ln(en)=1+\ln n,\quad n\ge 2$$ 
