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

$$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$$ when $$0

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.

• Note that this follows directly from math.stackexchange.com/a/1512630/42969: The left-hand side is equal to $\sin({\arctan}(a)-{\arctan}(b))$. – Martin R Mar 23 at 9:15
• Woow, it's really beautiful and brilliant. Thanks!! @MartinR – YerShane Mar 23 at 9:21
• Or you can take derivative (i.e. $\frac{d}{da}$) for Right subtract Left, and it is easy to see that is nonnegative. – Yu Ding Mar 23 at 9:21
• Actually let $f(x)=arctan x-arctan b-\frac{x-b}{\sqrt{1+x^2}\sqrt{1+b^2}}$, in the end you see $f'(x)\geq 0$ is reduced to $(1+bx)^2\leq (1+b^2)(1+x^2)$, which is equivalent to $2bx\leq b^2+x^2$. – Yu Ding Mar 23 at 9:38
• I'll add more details and try to re-open the question. @Matteo – YerShane Mar 24 at 1:28

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 $$\sin \alpha < \alpha$$.

• Your prove is so amazing that I ticked it. Thanks for your drawing and demonstration! – YerShane Mar 23 at 16:03
• @YerShane thanks for your appreciation! – Matteo Mar 23 at 16:05

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 and the integrated function is positive.

• Got it! Thanks~ – YerShane Mar 23 at 9:36
• @YerShane you can tick the answer if you believe it has been sufficient enough to answer your question! $\color{green}{\checkmark}$ – Mr Pie Mar 23 at 9:51
• I know, but I think the hint in the comment above is more useful...So should I tick this answer under this situation?@user477343 – YerShane Mar 23 at 9:57