Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$

Integrating by parts,

$$\int_a^b\frac{\sin x}x\,dx = \frac{\cos a}a-\frac{\cos b}b -\int_a^b\frac{\cos x}{x^2}\,dx.$$ Then apply the triangle inequality (twice) to get $$\bigg|\int_a^b\frac{\sin x}{x}\,dx\bigg|\leqslant \frac1a+\frac1b+\int_a^b\frac{1}{x^2}\,dx = \frac{2}{a},$$ since $\int_a^b\frac 1{x^2}\,dx = \frac1a-\frac 1b$.

\begin{align*} I &= \int_{a}^b \dfrac{\sin x}{x}dx \\ &= \dfrac{-\cos x}{x}|_{a}^b+ \int_{a}^b\dfrac{\cos x}{x^2}dx \\ &= \dfrac{\cos a}{a} - \dfrac{\cos b}{b}+ J \\ &\le \dfrac{1}{a}+\dfrac{1}{b}+ \int_{a}^b \dfrac{1}{x^2}dx \\ &= \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{b} \\ &= \dfrac{2}{a} \text{.} \end{align*} Also: $I \ge -\dfrac{1}{a} - \dfrac{1}{b}- \displaystyle \int_{a}^b \dfrac{1}{x^2}dx = -\dfrac{2}{a}$. Thus $|I| \le \dfrac{2}{a}$ as claimed.