Definite Integral Inequality

[jacks]

If \(f(x)\) is a positive function in \(\left[a,b\right]\). Then \( \displaystyle \left|\left(\int_{a}^{b}f(x)dx\right)\cdot \left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\right|\geq (b-a)^2\)

[Grigorios Kostakos]

Because \(f\) and \(\frac{1}{f}\) are positive functions, we have that \(\int_{a}^{b}f(x)\,dx\geqslant0\) and \(\int_{a}^{b}\frac{1}{f(x)}\,dx\geqslant0\). So \[\biggl({\int_{a}^{b}f(x)\,dx}\biggr)\,\biggl({\int_{a}^{b}\frac{1}{f(x)}\,dx}\biggr)\geqslant0\quad(1)\,.\] By the Cauchy-Schwarz inequality for the positive functions \(\sqrt{f}\) and \(\sqrt{\frac{1}{f}}\) we have that \[\displaystyle\biggl({\int_{a}^{b}{\sqrt{f(x)}\,}^2\,dx}\biggr)\,\biggl({\int_{a}^{b}{\sqrt{\frac{1}{f(x)}}\,}^2\,dx}\biggr)\geqslant\biggl({\int_{a}^{b}{\sqrt{f(x)}\,\sqrt{\frac{1}{f(x)}}}\,dx}\biggr)^2=\biggl({\int_{a}^{b}{1\,dx}}\biggr)^2=(b-a)^2\quad(2)\,.\] So
\begin{align*}
\displaystyle \left|\biggl({\int_{a}^{b}f(x)\,dx}\biggr)\,\biggl({\int_{a}^{b}\frac{1}{f(x)}\,dx}\biggr)\right|&\stackrel{(1)}{=}\biggl({\int_{a}^{b}f(x)\,dx}\biggr)\,\biggl({\int_{a}^{b}\frac{1}{f(x)}\,dx}\biggr)\\
&=\biggl({\int_{a}^{b}{\sqrt{f(x)}\,}^2\,dx\biggr)\,\biggl(\int_{a}^{b}{\sqrt{\frac{1}{f(x)}}\,}^2\,dx}\biggr)\\
&\stackrel{(2)}{\geqslant}(b-a)^2\,.\qquad\square
\end{align*}

[jacks]

Thanks Grigorios Kostakos