Prove $\ln\frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}$ for $0It 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.
 A: Hint: $f(x) = 1$, start with the LHS of the inequality
Hint 2: $g(x) = \frac1 x$
A: $\ln\left(\frac{p}{q} \right)\leq \sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}$, 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 $x\geq 1$.
A: Estimating $x \mapsto \tfrac{1}{x}$ on $[q,p]$ by a linear function that slopes down from $\tfrac{1}{q}$ to $\tfrac{1}{p}$ you get
$$
\log\frac{p}{q} = \int_q^p\frac{dx}{x} \leq \frac{1}{2}\left(\frac{1}{q}+\frac{1}{p}\right)(p-q) = \frac{1}{2}\frac{p^2-q^2}{pq}.
$$
Then also
$$
\log\frac{p}{q} = 2\log\frac{\sqrt{p}}{\sqrt{q}} \leq \frac{p-q}{\sqrt{pq}}.
$$
A: $f(x) = \frac{1}{x}, g(x) = 1$, 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 $$\ln\frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}$$
A: A proof I came up with
when this question was asked
in July 2019.
This does not use C-S.
We want to show that
$\log(p/q) \leq (p-q)/\sqrt(pq)
$
Let
$r = p/q$.
Since
$(p-q)/\sqrt(pq)
=\sqrt{p/q}-\sqrt{q/p}
$,
the inequality becomes
$\ln(r)
\le \sqrt{r}-\sqrt{1/r}
$.
Let $r = s^2$.
This becomes
$2\ln(s)
\le s-1/s
$.
This is true for $s=1$.
Let
$f(s)
=s-1/s-2\ln(s)
$.
$f(1) = 0$.
$f'(s)
=1+1/s^2-2/s
=(1-1/s)^2
\ge 0$.
Therefore
$f(s) \ge 0$
which is what we want.
