How to determine $\sup_{x \in \mathbb{R}} \left|\frac{\sin \left(x^{2}+1\right)}{x^{2}+1}\right|$

Question

It is clear (especially using a graphic calculator) that the sup of the function is the value of the function for $x=0$, but I want to prove it analytically.

First, because of Weierstrass, the sup of $f(x)=\frac{\sin (x^2+1)}{x^2+1}$ is equal to its max (the function is continuous everywhere on $\mathbb{R}$ and it's bounded since $f(x) \rightarrow 0$ for $x \rightarrow \pm \infty$), so I tried to do the derivative of the function to find the stationary points and determine the maximum. The problem is that the derivative is $$f'(x)=2x\dfrac{(x^2+1)\cos (x^2+1)-\sin (x^2+1)}{(x^2+1)^2}$$

So the stationary points are $x=0$ and the solutions of the equation $$\tan (x^2+1)=x^2+1 \quad (1).$$ Then, I thought that, since we should find the sup of the absolute value of $f$, we could define the sequence $\{|f(x_n)|\}_{n \in \mathbb{N}}$ where $\{x_n\}_{n \in \mathbb{N}}$ are the stationary points of the function such that $x_0=0$ and $x_n$ is a solution of the equation $(1)$ for every $n>0$ (in particular, we can take $\{x_n\} \subset [0,+\infty)$ whitout loss of generality because the function is even and so we have that $f(x_n)=f(-x_n)$ for every $n \in \mathbb{N}$) and verify that is a decreasing sequence, so that we have that $f(x_0)=f(0)$ is the sup of $|f|$.

The problem with this approach seems to be that there are no analytical solution to $(1)$. What do you suggest?

Answer 1

Hint

Let $a= \sqrt{\pi - 1}$. For $x \ge a$ we have

$$\vert f(x) \vert = \frac{\vert \sin(x^2+1) \vert}{x^2+1}\le \frac{1}{x^2+1} \le \frac{1}{a^2+1} = \frac{1}{\pi}.$$ Now use following facts:

1. $f$ is positive decreasing on $[0,a)$. This is essentially equivalent to prove that $0 \le x \le \tan x$ on $[0,\frac{\pi}{2})$.
2. $f(a)=0$.
3. $f(0) = \frac{\sin 1 } {1} \gt \frac{1}{\pi}.$

Which allow to conclude that $\sup\limits_{x \in \mathbb R} \vert f(x) \vert = f(0)$.