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I used LMVT to establish that $(\tan^{-1}(x)-\tan^{-1}(y))/(y-x) = 1/1+c^2$ After that I'm stuck and don't know how to establish the given inequality.

2 Answers 2

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Let $f(x) = \tan^{-1}x$, then $f'(x) = \frac{1}{1+x^2}$.

WLOG, suppose that $x < y$. By Mean Value Theorem, there exists $c \in (x,y)$ such that

$$\frac{f(x)-f(y)}{x-y} = f'(c)$$

$$\frac{\tan^{-1}x-\tan^{-1}y}{x-y} = \frac{1}{1+c^2} \le 1$$

Edit in response to @Mucciolo's comment

If we have $c \ne 0$, then we're done.

Thus, assume, $c = 0$. To avoid copying @zhw.'s solution, it's better to give an alternative solution by considering the strictly increasing function $h(t) = \tan t - t \;\forall t \in \left( -\frac\pi2,\frac\pi2 \right)$.

$h'(t) = \sec^2 t - 1 > 0\;\forall |t| \in \left( 0,\frac\pi2 \right)$ so it's strictly increasing on $\left( -\frac\pi2,0 \right)$ and $\left( 0,\frac\pi2 \right)$.

Let $s < 0 < t$. Then there exists $\xi,\zeta$ with $s < \xi < 0 < \zeta < t$ such that \begin{alignat}{2} \frac{h(s) - h(0)}{s-0} &= h'(\xi) > 0 &\quad \frac{h(t) - h(0)}{t-0} &= h'(\zeta) > 0 \\ h(s) - h(0) &< 0 &\quad h(t) - h(0) &> 0 \end{alignat} So we have $h(s) 0$. This shows that $h$ is strictly increasing, and thus injective.

Let $s = \tan^{-1} x$ and $t = \tan^{-1} y$.

\begin{align} \tan^{-1}x-\tan^{-1}y &= x-y \\ s - t &= \tan s - \tan t \\ \tan s - s &= \tan t - t \\ h(s) &= h(t) \\ s &= t \quad (h \text{ is injective}) \\ \tan s &= \tan t \\ x &= y \quad \text{(contradicts $x < y$)} \end{align}

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    The inequality $ \frac{1}{1+c^2}<1 $ holds provided that $ c \neq 0 $. So the case where $ c = 0 $ still need to be addressed. – 2017-02-25
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    @Mucciolo Thanks for your comment (and I guess, downvote). I've edited my answer in order to address the problem that you had raised. Though the trick ("zero-in-the-middle") is the same as zhw.'s answer, I tried not copying his solution. – 2017-02-25
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Suppose WLOG $x

$$\tag 1 f(y) - f(x) = f(y)-f(0) + f(0) -f(x) = f'(c)(y-0) + f'(d)(0-x),$$

where $0