Logarithmic inequality January 26, 2021 Tolaso Let . Prove that Solution Let and . Thus, Thus, The result follows. Read more They might interest you:Parametric improper integralOn the sum of inverse binomialAn Euler's gamma constant limitAn Euler squared sumA zeta seriesBessel seriesExistence of cPell - Lucas seriesAn infinite product
. Prove that![\[\ln \frac{p}{q} \leq \frac{p-q}{\sqrt{pq}}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-76e0f354a6eef62a56cc0764b9e7d9c6_l3.png "Rendered by QuickLaTeX.com")
and 
. Thus,![\[\left(\int_{q}^{p} \frac{1}{x} \, \mathrm{d}x \right)^2 \leq \int_{q}^{p} \frac{1}{x^2}\, \mathrm{d}x \int_{q}^{p} 1 \, \mathrm{d}x\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-8b3aab53614803892914f9bb219a4de5_l3.png "Rendered by QuickLaTeX.com")
. Prove that and . Thus,