<s>[INST] Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!
 [/INST] Inequality that is another form of a well-known inequalities:
For $0<y<x$
$$ \sqrt{xy}<\frac{x-y}{\ln x-\ln y} < \frac{x+y}{2}. (1)$$
For $a>b$ inequality of enunciation transform so:
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(a-b)(\ln a-\ln b)<=>$$
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})(\ln a-\ln b)<=>$$
$$\sqrt{a} - \sqrt{b} < \frac{1}{2}(\sqrt{a}+\sqrt{b})(\ln \sqrt{a}-\ln \sqrt{b})<=>$$
$$\frac{\sqrt{a} - \sqrt{b} }{\ln \sqrt{a}-\ln \sqrt{b}}< \frac{\sqrt{a}+\sqrt{b}}{2}$$
which is obtained from (1) for $x=\sqrt{a}, y=\sqrt{b}.$
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