<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] Alternative solution:
When $n = 2, 3, 4$, the inequality is verified directly.
In the following, assume that $n\ge 5$.
Let
$$I_n = \int_0^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x.$$
We have
\begin{align}
I_n &= \underbrace{\int_0^{\pi/n} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,1}}
+ \underbrace{\int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,2}}.
\end{align}
First, we have
\begin{align}
I_{n,1} &\le \int_0^{\pi/n} \frac{x}{n^2}(\sin nx)^4 \frac{1}{x^4} \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4\mathrm{d} x \\
&= \frac{1}{n^2}\left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi/n} \frac{(\sin nx)^4}{x^3} \mathrm{d} x \\
&= \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y\\
&\le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
\end{align}
where we have used: i) $\frac{\sin x}{x} \ge \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$
on $0 \le x \le \frac{\pi}{n}$; ii) the substitution $y = nx$; iii) $\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$ is non-increasing for $n\ge 2$.
Second, we have
\begin{align}
I_{n, 2} &= \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x\\
&\le \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\pi}{2x}\right)^4\mathrm{d} x \\
&= -\frac{\pi^2}{8n^2} + \frac{\pi^2}{32}\\
&\le \frac{\pi^2}{32}
\end{align}
where we have used $\sin x \ge \frac{2}{\pi}x$ for $0 \le x \le \frac{\pi}{2}$.
Thus, we have
$$I_n \le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
+ \frac{\pi^2}{32} < \frac{\pi^2}{8}.$$
We are done.
</s>