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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.

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In order for your solution above to be complete I think you need to verify that $F(q)\geq0$? – Henrik Mar 19 '12 at 15:50
up vote 11 down vote accepted

$f(x) = \frac{1}{x}, g(x) = 1$, 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 $$\ln\frac{p}{q}\leq \frac{p-q}{\sqrt{pq}}$$

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

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

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any more detail? – 89085731 Mar 19 '12 at 12:39
@Gingerjin, check my next hint (which should now give it all away) – davin Mar 19 '12 at 12:44

$\ln\left(\frac{p}{q} \right)\leq \sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}$, 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 $x\geq 1$.

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read my explanation. – 89085731 Mar 19 '12 at 12:38

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

$$ \log\frac{p}{q} = 2\log\frac{\sqrt{p}}{\sqrt{q}} \leq \frac{p-q}{\sqrt{pq}}. $$

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