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I have a problem here that asks to show that the function $ f: [0,\infty) \to \mathbb{R} $ defined by $$ f(x) \stackrel{\text{df}}{=} \begin{cases} \dfrac{1}{x} \left( 1 + \dfrac{x^{2}}{4} \right) \tan^{-1} \! \left( \dfrac{x}{2} \right), \qquad \text{if $ x > 0 $}; \\ \dfrac{1}{2}, \qquad \text{if $ x = 0 $} \end{cases} $$ is strictly increasing without the use of graphs. Now, I know how to do the problem, but I believe that my solution is overly complicated.


Let me explain my solution:

Observe that $ f $ is continuous on $ [0,\infty) $ and differentiable on $ (0,\infty) $. Its derivative is given by $$ \forall x \in (0,\infty): \quad f'(x) = \frac{(x^{2} - 4) \tan^{-1} \! \left( \dfrac{x}{2} \right) + 2 x} {4 x^{2}}. $$ It suffices to prove that $ f' > 0 $ on $ (0,\infty) $. As $ f' $ is continuous and $ f' \! \left( \dfrac{\pi}{2} \right) > 0 $, we only need to show that the equation $$ (x^{2} - 4) \tan^{-1} \! \left( \frac{x}{2} \right) + 2 x = 0 $$ has no solutions in $ (0,\infty) $. Obviously, $ x = 2 $ is not a solution, so it is enough to prove that $$ (\diamond) \qquad \tan^{-1} \! \left( \frac{x}{2} \right) + \frac{2 x}{x^{2} - 4} = 0 $$ has no solutions in $ (0,2) \cup (2,\infty) $. In fact, we can rule out the interval $ (2,\infty) $ because then the left-hand side of $ (\diamond) $ would be positive. We shall thus focus our attention on $ (0,2) $.

Define a function $ g: (0,2) \to \mathbb{R} $ by $$ \forall x \in (0,2): \quad g(x) \stackrel{\text{df}}{=} \tan^{-1} \! \left( \frac{x}{2} \right) + \frac{2 x}{x^{2} - 4}. $$ Then $ g $ is differentiable on $ (0,2) $ and $$ \forall x \in (0,2): \quad g'(x) = \frac{2}{x^{2} + 4} - \frac{2 (x^{2} + 4)}{(x^{2} - 4)^{2}} = - \frac{32 x^{2}}{(x - 2)^{2} (x + 2)^{2} (x^{2} + 4)} < 0. $$ Hence, $ g $ is strictly decreasing on $ (0,2) $, and as $ \displaystyle \lim_{x \to 0^{+}} g(x) = 0 $, it follows that $ g $ has no roots in $ (0,2) $, or equivalently, $ (\diamond) $ has no solutions in $ (0,2) $.

Conclusion: $ f $ is strictly increasing on $ [0,\infty) $.


Question. Can a simpler but rigorous argument be found?

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Compose with the strictly increasing function $x=2\tan(z/2)$, $z\in(0,\pi)$. Such composition doesn't change the monotonicity. We get

$$f(2\tan(z/2))=\frac{(1+\tan^2(z/2))z/2}{2\tan(z/2)}=\frac{z}{2\sin(z)}$$

And now $g(z)=\frac{\sin(z)}{z}$ is easy(er) to study in $(0,\pi)$.

For this we can proceed similar to what you did, by looking at the derivative.

$$g'(z)=\cos(z)\frac{z-\tan(z)}{z^2}$$

Now $z-\tan(z)\leq 0$, for $z\in(0,\pi/2]$, $z-\tan(z)\geq0$, for $z\in[\pi/2,\pi)$, while $\cos(z)\geq0$ and $\cos(z)\leq0$ in the same corresponding intervals. This can be seen in many ways, even geometrically, from the definition of $\tan(z)$ and $\cos(z)$:

enter image description here

Therefore, $g'(z)\leq0$ for $z\in(0,\pi)$ and it follows that $g$ is decreasing, and then that $f$ is increasing.

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  • $\begingroup$ Thank you very much, Alamos. $\endgroup$ – Transcendental Apr 21 '15 at 22:12

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