Inequality

\hspace{-16}$Here $\bf{f}$ and $\bf{\frac{1}{f}}$ are positive function in $\bf{[a,b]}$.\\\\\\ So $\bf{\int_{a}^{b}f(x)dx\geq 0}$ and $\bf{\int_{a}^{b}\frac{1}{f(x)}dx\geq 0}.$\\\\\\ So $\bf{\left(\int_{a}^{b}f(x)dx\right).\left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\geq 0}$\\\\\\ So By Using Cauchy-Schwarz Inequality for $\bf{\sqrt{f}}$ and $\bf{\sqrt{\frac{1}{f}}.}$\\\\\\ $\bf{\left(\int_{a}^{b}\left(\sqrt{f(x)}\right)^2\right)\cdot \left(\int_{a}^{b}\left(\sqrt{\frac{1}{f(x)}}\right)^2\right)\geq \left(\int_{a}^{b}\sqrt{f(x)}.\sqrt{\frac{1}{f(x)}}\right)^2=\left(\int_{a}^{b}1.dx\right)^2 = (b-a)^2}$\\\\\\ So $\bf{\left|\left(\int_{a}^{b}f(x)dx\right)\cdot \left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\right|= \left(\int_{a}^{b}f(x)dx\right)\cdot \left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\geq (b-a)^2}$\\\\\\ So $\boxed{\boxed{\bf{\left|\left(\int_{a}^{b}f(x)dx\right)\cdot \left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\right|\geq (b-a)^2}}}$

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