试证: $$\bex 0<q<p\ra \ln \frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}. \eex$$
证明: $$\bex \sex{\ln \frac{p}{q}}^2 =\sex{\int_q^p \frac{1}{x}\rd x}^2 \leq \int_q^p1^2\rd x \cdot \int_q^p \frac{1}{x^2}\rd x =(p-q)\sex{\frac{1}{q}-\frac{1}{p}} =\frac{(p-q)^2}{pq}. \eex$$
