Given $1\leq a<b\leq2$, Prove $\frac{log(b)-log(a)}{b^2-a^2}<\frac{1}{2}$

Question

A question that was in my calculus 1 test, I remember the question from the mind, I'm pretty sure I didn't miss anything:
Given $1\leq a<b\leq2$, Prove that $$\frac{log(b)-log(a)}{b^2-a^2}<\frac{1}{2}$$
I would like to know the correct form to solve these kinds of questions.
What I did is that I showed the "Edge Cases". First is when $b=max[1,2]$ and $a\to max[1,2]$, so actually $b=2 , a \to 2$. We get:
$$\frac{log(2)-log(a)}{2^2-a^2}=L'Hoptial's \frac{0}{0}=\frac{-\frac{1}{a}}{-2a}=\frac{1}{2a^2}\to\frac{1}{8}<\frac{1}{2}$$ I did the same for $a=1 , b\to1$ where I got the term is $\to{\frac{1}{2}}^{-}$ which is $<\frac{1}{2}$. Also $a=1 , b=2$, And it works. Then I said that these cases cover all other cases when $1\leq a<b\leq2$. Is it a correct way to show that? What's a proper way to prove that? Thanks.

Answer 1

"Calculus" is the key word. :)

By the mean value theorem: there is $c\in(a,b)$ such that $$ \frac{\log(b)-\log(a)}{b-a}\frac{1}{b+a}=\frac{1}{c}\frac{1}{b+a}<\frac{1}{1}\frac{1}{1+1}=\frac{1}{2}. $$ The inequality is because $c>a\geq 1$ and $b+a>a+a\geq 1+1$.

Answer 2

Note that the function $f(x) =e^x /x$ is monotonically increasing for $x \geq 1$, in particular \begin{eqnarray*} \frac{e^{a^2}}{a^2} < \frac{e^{b^2}}{b^2} \end{eqnarray*} which can be rearranged to give the required inequality.

Answer 3

General outline:

Let $f(x)=2\log(x)-x^2$ on $[1,2]$

Note that $f'(x)=\frac{2}{x}-2x=2\left(\frac{1}{x}-x\right)\leq 0$ on this interval, so the function decreases.

In particular, if $1<a<b\leq 2$, then $f(a)>f(b)$.

Which implies

$$2\log(a)-a^2> 2\log(b)-b^2$$

which leads to the required result.