Prove $\ln\frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}$ for $0<q\leq p$

$f(x) = \frac{1}{x}, g(x) = 1$, and therefore

$$ \left(\int_{q}^{p} \frac{1}{x} dx \right)^2 \leq \int_{q}^{p} \frac{1}{x^2}dx \int_{q}^{p} 1 dx$$

which means

$$ \begin{align*} (ln(p)-ln(q))^2 &\leq \left(-\frac{1}{p}+\frac{1}{q}\right) \times (p-q)\\ &=\frac{(p-q)^2}{pq}\\ \end{align*} $$

Therefore $$\ln\frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}$$

Hint: $f(x) = 1$, start with the LHS of the inequality

Hint 2: $g(x) = \frac1 x$

$\ln\left(\frac{p}{q} \right)\leq \sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}$, putting $x=\frac{p}{q}(\geq 1$ because $q\leq p)$ we have: $\ln x\leq \frac{x-1}{\sqrt x}$ and this relation is true in every interval $[1,M>0)$ and since $\lim_{x\rightarrow \infty} \frac{\ln x}{x}=0$ then it is true for every $x\geq 1$.