Can we show $x>\ln(1+x^2)$ for all $x>0$?

Can we show $x>\ln(1+x^2)$ for all $x>0$?

Question

Given that $y=x-\ln(1+x^2),$ show that $\frac{dy}{dx}\geq 0\ \text{for all values of }x\text{.}$

Deduce that, for all $x>0$, $x>\ln(1+x^2).$

Solution

We can compute

$\dfrac{dy}{dx}$

using the chain rule:

$$\begin{align*} \frac{dy}{dx} &= 1-\frac{1}{1+x^2}\times\frac{d}{dx}(1+x^2)\\ &= 1-\frac{2x}{1+x^2}\\ &= \frac{1+x^2-2x}{1+x^2}\\ &= \frac{(x-1)^2}{1+x^2}. \end{align*}$$

We know that $(x-1)^2 \geq 0$ and $1+x^2\geq 1$ for all values of $x$, so $\dfrac{dy}{dx} \geq 0$ for all values of $x$.

We know that $\dfrac{dy}{dx}\geq 0$ for all values of $x$, so the function $x-\ln(1+x^2)$ is an increasing function.

Substituting we find that when

$x = 0$

,

$y=0$

and

$\dfrac{dy}{dx} = 1$

. This means that for any value of

$x>0$

,

$y$

must be greater than zero. So

$$\begin{align*} x &- \ln(1+x^2) > 0\\ \implies\quad x &> \ln(1+x^2) \end{align*}$$

for all $x>0$ as required.