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

It is a question in one problem book:

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

Actually I already solved it: Define $F(x)=\frac{x-q}{\sqrt{xq}}-\ln x+\ln q$, then $F'(x)\geq0$ when $x\geq q$.

However,the problem book gives a hint to use Schwarz inequality $$\left(\int_a^b f(x)g(x)dx\right)^2\leq \int_a^b f^{2}(x)dx\cdot\int_a^bg^2(x)dx$$ I don't know how to use it.

$f(x) = \frac{1}{x}, g(x) = 1$, 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: $g(x) = \frac1 x$