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.

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.
