Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like 
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me some hints/guides here?
Thanks.

$$
\begin{array}{rcl}
\ln(1 + x^2) & \le & x^2 \\ 
e^{\ln(1 + x^2)} & \le & e^{x^2} \\ 
1 + x^2 & \le & e^{x^2} \\ 
1 + x^2 & \le & 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \\  {}
0 & \le & \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \dots \\ {}
\end{array}
$$
Which is true for all real $x$.
