So I'm having trouble trying to solve this problem:
Let $a$ and $b$ be two real numbers such that $a \geqslant b \geqslant 1$. Using the Mean value Theorem prove that
$$\ln\left(\frac{a}{b}\right)\leqslant a-b$$
I honestly don't even know how to start, so I was wondering if someone could please lend me a hand? Thank you very much.
Best Answer
If $a=b$, we are done. Otherwise, by the MVT, there is some $c\in(b,a)$ such that $$ \frac{\log(a)-\log(b)}{a-b}=\frac{1}{c}\leq 1 $$ where the last inequality is because $c\geq b\geq 1$.