How to prove that $\frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$ when $0$$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$$ when $0<b<a$
This might relate to the mean value theorem, but I just can't prove it.

This question was put on hold as off-topic, I couldn't understand the reason why people voted to close it so I add more details and try to re-open it.
After I learned the course of the mean value theorem, my teacher asked us to prove that $$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b} < a-b $$
I found by using the equation $\arctan a - \arctan b = \frac{1}{1+\xi^2}(a-b)$ I could easily prove that $$\frac{a-b}{1+a^2} < \arctan{a}- \arctan{b} < a-b $$
the right inequality related to several questions on StackExchange so I omitted it. 
I also tried to use the inequality $\frac{x}{1+x^2} < \arctan x < x$ combined with the equation $\arctan x - \arctan y = \arctan{\frac{x-y}{1+xy}}$ to prove this question but I failed again.
So I went to check if this question is correctly written and my teacher said "Yes, nothing wrong with it".
I found the comment of @YuDing is more useful than the answer of @AdamLatosiński so I didn't tick the answer and I also couldn't tick the comment because it's just a comment.
The comment of @MartinR and the answer of @Matteo provided us a great perspective to solve the question. I ticked the answer for the reason that I couldn't tick a comment. Maybe because the answer is on a purely geometrical perspective so my question is off-topic?
I hope OP could re-open this question because I really appreciate everyone's efforts here. Thank you all.
 A: I don't add much with respect to the comment by Martin: just a purely geometrical perspective on the situation. Consider the figure below, where $\triangle ABC$ is right-angled and $\overline{AB} =1$, $\overline{BC} = b$, and $\overline{BD} = a$. From $C$ draw the line perpendicular to $AD$, that intersects $AD$ in $E$.

By definition you have
$$\angle CAB = \arctan b$$
and
$$\angle DAB = \arctan a,$$
so that
$$\angle DAC = \arctan a- \arctan b.$$
Now use the fact that $\triangle CDE \sim \triangle ABD$ to determine
$$\overline{CE} = \frac{a-b}{\sqrt{1+a^2}}.$$
Thus, as already noted in the above mentioned comment
$$\sin \angle DAC = \frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}},$$
and the inequality follows from $\sin \alpha < \alpha$.
A: You can write right side as
$$ \arctan a - \arctan b = \int_b^a \frac{1}{1+x^2}dx $$
and left side as
$$ \frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} = \int_b^a \frac{\partial}{\partial x}\left(\frac{x-b}{\sqrt{1+x^2}\sqrt{1+b^2}}\right) dx = \int_b^a \frac{xb+1}{(1+x^2)^\frac32\sqrt{1+b^2}} dx$$
Therefore the inequality you want to prove can be written as
\begin{align} 0 &< \int_b^a \left(\frac{1}{1+x^2}-\frac{xb+1}{(1+x^2)^\frac32\sqrt{1+b^2}}\right) dx = \\
&\quad = \int_b^a \frac{\sqrt{1+x^2}\sqrt{1+b^2}-(xb+1)}{(1+x^2)^\frac32\sqrt{1+b^2}} dx = \\
&\quad = \int_b^a \frac{(1+x^2)(1+b^2)-(xb+1)^2}{(1+x^2)^\frac32\sqrt{1+b^2}\big(\sqrt{1+x^2}\sqrt{1+b^2}+(xb+1)\big)} dx = \\
&\quad = \int_b^a \frac{(x-b)^2}{(1+x^2)^\frac32\sqrt{1+b^2}\big(\sqrt{1+x^2}\sqrt{1+b^2}+1+xb\big)} dx\end{align}
which is true because $b<a$ and the integrated function is positive.
