How to prove that $\ln(1+x^2)<x$ , given that $x>0$

Question

Given that $x>0$, how to prove that $\ln(1+x^2)<x$?

I have been thinking about Taylor series, but didn't know how to do it. any suggestions?

Answer 1

Consider the function $f(x)=\ln(1+x^2)-x$; then $f(0)=0$, while $$ f'(x)=\frac{2x}{1+x^2}-1=-\frac{(1-x)^2}{1+x^2} $$ which is negative except for $x=1$. Thus $f$ is strictly decreasing.

Answer 2

Assuming $x>0$...

$\log(1+x^2) - x = \log(1+x^2)- \log e^x = \log ( {1+x^2 \over e^x } ) $.

Since $e^x > 1+ x^2$, we see that ${1+x^2 \over e^x } < 1$ hence $\log(1+x^2) - x < 0$.

Answer 3

Since $e^x$ is monotone increasing, your inequality is equivalent to

$$1 + x^2 < e^x,$$

applying $\exp(x)$ to each side of the original inequality.