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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.

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 $F(q)\geq0$?2012-03-19

5 Answers 5

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$f(x) = \frac{1}{x}, g(x) = 1$, 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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    @Gingerjin, check my next hint (which should now give it all away)2012-03-19
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$\ln\left(\frac{p}{q} \right)\leq \sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}$, 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.2012-03-19
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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}}. $