# Can a simple but rigorous argument be found to prove that this function is strictly increasing?

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?

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 $\cos(z)$:

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

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