Prove $\left|\frac1{x^2}\int_0^x\sin\frac1tdt\right|<1$

Question

I want to prove $$\left|\frac1{x^2}\int_0^x\sin\frac1tdt\right|<1,\ x\ne0$$ I set $$f(x)=\left|\frac1{x^2}\int_0^x\sin\frac1tdt\right|$$ And I found that $$f\left(\frac1x\right)=\left|x^2\operatorname{Ci}(|x|)-x\sin x\right|$$ As $f(x)=f(-x)$, I can assume $x>0$.
I noticed that $$x^2\operatorname{Ci}(x)-x\sin x\sim\cos x,\ x\to\infty$$ So $\limsup\limits_{x\to0}f(x)=1$.
But how can I prove $x^2\operatorname{Ci}(x)-x\sin x<1$?