Proving that $\forall x \geq 0:\frac{x}{\sqrt{x^2 + 1}} \leq \arctan x$

Question

I have the following statement:

Prove that $\forall x \geq 0: \frac{x}{\sqrt{x^2 + 1}} \leq \arctan x$.

My development was:

I use the mean value theorem in the interval $[0, x]$ and I got:

$\exists c \in [0,x]: f'(c) = \frac{\arctan x}{x} \implies \frac{x}{1+c^2} = \arctan x$

Since $c \in [0,x]$ i have : $0\leq c \leq x$, doing some work with inequalities I got:

$$x \geq \arctan(x) \geq \frac{x}{x^2 +1}$$

From here i don't know what I can do to get the desired result.

I had thought of some kind of transitivity but I couldn't think of anything interesting.

From the graph I can see that $\frac{x}{\sqrt{x^2 + 1}}$ is a graph than approximates $\arctan(x)$ better than$\frac{x}{x^2 + 1}$.

I'll appreciate any hint.

With $\forall x$ you probably mean for all non-negative $x$ I guess? Because otherwise the statement is false. — SampleTime, Jul 12 '20 at 17:55
Hint: $\arctan(x)=\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$ — robjohn, Jul 12 '20 at 21:24

score 3 · Answer 1 · answered Jul 12 '20 at 17:48

3

Replace $x$ by $\tan(t)$. Then the statement becomes obvious (and well known).

answered Jul 12 '20 at 17:48

markvs

Well, but $\tan$ is not positive for all $x \geq 0$, so what you mean? – ESCM Jul 12 '20 at 20:37

1 Answers1