how to prove this inequlity $\ln(\frac{p}{q}) \le \frac{p-q}{\sqrt{pq}} (0 < q \le p)$ by means of integral inequalities I can solve it by introducing a function, but I need to prove it by applying integral inequalities such as the holder inequality or the schwarz inequality.

You probably can recognize this is the equivalent to prove
$$
f(x) = \ln(x)\le \sqrt{x} - \frac{1}{\sqrt{x}} = g(x) \quad \text{when} x \ge 1
$$
Easy to see $f(1) = g(1)= 0$
We also have
$$
g'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x}^3} = \frac{1}{2}\left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}^3}\right) \ge \sqrt{ \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}^3}} = \frac{1}{x} = f'(x)
$$
Thus $f(x)\le g(x)$ when $x \ge 1$ as a result of integral inequalities.
