Is it true that $\sqrt{ab}\le \frac{a-b}{\ln a - \ln b}$ for any $a\neq b>0$? [closed]

Question

• Is it true that $\sqrt{ab}\le \frac{a-b}{\ln a - \ln b}$ for any $a\neq b>0$?

If so, any thoughts on how to prove this?

Answer 1

The assertion is true. When $a>b>0$, it follows by taking $x:=\frac ab$ in the inequality $$ \log x\le \sqrt x - \frac1{\sqrt x}\qquad\text{for all $x\ge 1$} $$ For $b>a$, just flip the roles of $a$ and $b$.

Answer 2

The logs only make sense if $a,b>0$. Let $a=e^{B+h},b=e^B$; without loss of generality $h>0$, i.e. $a>b$. Then $$ \frac{a-b}{\ln a-\ln b} = \frac{e^{B+h}-e^B}{h}; $$you will recognize this as the slope of the secant line of $e^x$ at the points $x=B,x=B+h$. On the other hand, $$ \sqrt{ab}= \sqrt{e^{B+h}e^B} = e^{B+h/2} $$Divide by $e^B$ and substitute $h\to 2h$: $$ e^{h/2} \leq \frac{e^h-1}{h} $$ $$ 2he^{h} \leq {e^{2h}-1} $$This last inequality is true by Bernoulli's inequality: we have equality at $h=0$ and taking derivatives we have $$ 2e^h(1+h) \leq 2 e^{2h};\qquad 1+h \leq e^h $$