Proof First, we prove the lemma
$$|\sin (x-1)|+|\sin x|+|\sin (x+1)| \geqslant 2 \sin 1$$

Let the function $f(x)=|\sin (x-1)|+|\sin x|+|\sin (x+1)|$. It is easy to see that $f(x)$ is a periodic function with period $\pi$, and $f(\pi-x)=f(x)$, so we only need to prove that it holds for $0 \leqslant x \leqslant \frac{\pi}{2}$.
(1) When $1 \leqslant x \leqslant \frac{\pi}{2}$, we have
$$\begin{aligned} f(x)= & \sin (x-1)+\sin x+\sin (x-1)= \\ & \sin x \cos 1-\cos x \sin 1+\sin x+\sin x \cos 1+\cos x \sin 1= \\ & (2 \cos 1+1) \sin x \geqslant\left(2 \cos \frac{\pi}{3}+1\right) \sin x=2 \sin x \geqslant 2 \sin 1 \end{aligned}$$
(2) For $0 \leqslant x \leqslant 1$, we have
$$f(x)=\sin (1-x)+\sin x+\sin (1+x)=2 \sin 1 \cos x+\sin x$$

Let $\phi=\arctan (2 \sin 1)$, it is easy to see that $\frac{\pi}{4} \\
\frac{2}{3} \sin 1\left(\frac{1}{1+\frac{\pi}{3}}+\frac{1}{2+\frac{\pi}{3}}+\frac{1}{3+\frac{\pi}{3}}+\sum_{k=4}^{149} \int_{k}^{k+1} \frac{\mathrm{d} x}{k+\frac{\pi}{3}}\right)= \\
\frac{2}{3} \sin 1\left(\frac{1}{1+\frac{\pi}{3}}+\frac{1}{2+\frac{\pi}{3}}+\frac{1}{3+\frac{\pi}{3}}+\int_{4}^{150} \frac{\mathrm{d} x}{k+\frac{\pi}{3}}\right)= \\
\frac{2}{3} \sin 1\left(\frac{1}{1+\frac{\pi}{3}}+\frac{1}{2+\frac{\pi}{3}}+\frac{1}{3+\frac{\pi}{3}}+\ln \left(150+\frac{\pi}{3}\right)-\ln \left(4+\frac{\pi}{3}\right)\right)>
\end{array}$$$$
$2.5033>\frac{5}{2}$
Thus, the original inequality is proved.