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

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    $\begingroup$ Note that this follows directly from math.stackexchange.com/a/1512630/42969: The left-hand side is equal to $\sin({\arctan}(a)-{\arctan}(b))$. $\endgroup$ – Martin R Mar 23 '19 at 9:15
  • $\begingroup$ Woow, it's really beautiful and brilliant. Thanks!! @MartinR $\endgroup$ – YerShane Mar 23 '19 at 9:21
  • $\begingroup$ Or you can take derivative (i.e. $\frac{d}{da}$) for Right subtract Left, and it is easy to see that is nonnegative. $\endgroup$ – Yuval Mar 23 '19 at 9:21
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    $\begingroup$ 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 $2bx\leq b^2+x^2$. $\endgroup$ – Yuval Mar 23 '19 at 9:38
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    $\begingroup$ I'll add more details and try to re-open the question. @Matteo $\endgroup$ – YerShane Mar 24 '19 at 1:28
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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 $E$.

enter image description here

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 $\sin \alpha < \alpha$.

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  • $\begingroup$ Your prove is so amazing that I ticked it. Thanks for your drawing and demonstration! $\endgroup$ – YerShane Mar 23 '19 at 16:03
  • $\begingroup$ @YerShane thanks for your appreciation! $\endgroup$ – dfnu Mar 23 '19 at 16:05
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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.

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    $\begingroup$ Got it! Thanks~ $\endgroup$ – YerShane Mar 23 '19 at 9:36
  • $\begingroup$ @YerShane you can tick the answer if you believe it has been sufficient enough to answer your question! $\color{green}{\checkmark}$ $\endgroup$ – Mr Pie Mar 23 '19 at 9:51
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    $\begingroup$ I know, but I think the hint in the comment above is more useful...So should I tick this answer under this situation?@user477343 $\endgroup$ – YerShane Mar 23 '19 at 9:57

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