Prove $\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}$ 
Prove $$\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}.$$

My attempt:
\begin{align}
\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x & =\sum_{k=1}^n \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x\\
& \leq\sum_{k=1}^n \left({\pi\over 2}\right)^4 \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}\left({\sin^4nx\over x^3}\right)\mathrm{d}x \quad (\text{use } \sin x\geq {2\over \pi}x ) \tag{1}\label{1}\\
&= \left({\pi\over 2}\right)^4 n^2\sum_{k=1}^n  \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x \quad (\text{use } x\to {x\over n}).\\
\end{align}
Is my direction right? If right, how can I prove the following $$\sum_{k=1}^n  \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x\leq\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x \leq {2\over \pi^2}. $$
I use Mathematica to calculate the integral $\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x\simeq 0.7>{2\over\pi^2}$, hence my process (\ref{1}) seems to be wrong.

The term $\left(\frac{\sin nx}{\sin x}\right)^4$ is associated with the Jackson kernel.
Your inequality is indeed just a minor variation on Lemma 0.5 in the linked notes, and it can be proved through the same technique: expand $|x|$ as a Fourier cosine series over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, do the same for $\left(\frac{\sin nx}{\sin x}\right)^4$, then apply orthogonality/Bessel's inequality.
