How to prove that $\frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$ when $0<b<a$?

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 $\sin \alpha < \alpha$.

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.