<s>[INST] How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$
But using mathematica I found this inequality can still be improved.
And after calculating some terms I found it seems that when $n\geq 2$ we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$
But I cannot prove this.So is there any method to improve my result?Any help will be thanked.
 [/INST] Let $ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $
$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$
$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x} dx=2\int_{0}^{\frac{\pi}{2}} \cos(2n-2)x dx=0 $$
$$ a_{n}-2a_{n-1}+a_{n-2}=0 $$
So $ a_{n} $ is an arithmetic sequence,$ a_{n}=a+bn,a_{0}=0,a_{1}=\frac{\pi}{2} $
We get
$$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n\pi}{2} $$
Thus
$$ \int_{0}^{\frac{\pi}{2}}x \frac{\sin^4 nx}{\sin^4 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \frac{\left| \sin nx \right |}{\sin x} \left| \sin nx \right | \frac{\sin^2 nx}{\sin^2 x} dx ＜ \int_{0}^{\frac{\pi}{2}} \frac{\pi}{2} \times n \times 1 \times \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n^2 \pi^2}{4} $$
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