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.
 A: $$
\begin{array}{rcl}
\ln(1 + x^2) & \le & x^2 \\ 
e^{\ln(1 + x^2)} & \le & e^{x^2} \\ 
1 + x^2 & \le & e^{x^2} \\ 
1 + x^2 & \le & 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \\  {}
0 & \le & \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \dots \\ {}
\end{array}
$$
Which is true for all real $x$.
A: Yes, but it is a nuisance to use the Taylor expansion of $\log(1+x^2)$. Instead, let $t=x^2$, and look at what the Taylor expansion of $\log(1+t)$ tells you for non-negative  $t$.
Use the Lagrange form of the remainder. The error when you truncate the expansion of $f(t)$ at the linear term is equal to $\frac{1}{2!}f''(\xi)t^2$, where $\xi$ is a number between $0$ and $t$. Since the second derivative in our case is negative, truncating at the $t$ term gives us a negative error term, meaning that the first term $t$ overestimates $\log(1+t)$.
Or else if you are familiar with alternating series, you can get the same result.
There are plenty of other ways to prove the inequality. For example, let $g(t)=\log(1+t)-t$. We have $g(0)=0$, and $g'(t)=\frac{1}{1+t}-1$. So for $t\gt 0$, the function $g(t)$ is decreasing, and therefore $g(t)\lt 0$ if $t\gt 0$. That shows $\log(1+t)\lt t$ if $t\gt 0$.
A: Sure.
$$\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots$$
In particular, the terms alternate sign. If we subtract $x^2$, we are left with
$$\ln(1 + x^2) - x^2 = -\frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \ldots$$
The first term of the sum is negative, and since the sum is a strictly alternating sum whose terms go to $0$ (since $|x| < 1$), the error between a sum of the first $n$ terms and the actual value is bounded by the $n+1$st term. Here, we only need to note that the first term is negative and the second term is smaller in absolute value - thus we conclude that 
$$\ln(1 + x^2) - x^2 \leq 0$$
with equality only when $x = 0$.
A: As $1+t^2\ge 1$
$$\ln{\left(1+x^2\right)}=\int_{0}^{x}{\frac{2t}{1+t^2}dt}\le\int_{0}^{x}{\frac{2t}{1}dt}\le  x^2$$
