Given $1\leq aA 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.

"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 $b+a>a+a\geq 1+1$.
