Proving that $e^x>1+x^2$ I am trying to prove that $e^x>1+x^2$ for any $x>0$ for my homework assignment.
However I have run into trouble doing this. I was trying to probe that $\ln {{e}^{x}}>\ln (1+{{x}^{2}})$ is true for $x>0$ and then that would mean that  $e^x>1+x^2$ is true because $\ln x$ is a monotone rising function.
However I have come to the following conclusion$$\frac{{{x}^{2}}}{1+{{x}^{2}}}\le \ln (1+{{x}^{2}})\le {{x}^{2}}$$
which means $x\le \frac{{{x}^{2}}}{1+{{x}^{2}}}$ must be true. but it is not.
I am wondering where I made a mistake here - Or perhaps where I made many mistakes?
Maybe there is a much better why to solve this question also?
Thanks a lot :)
 A: Here's how I would proceed. Consider the function $f(x) = e^x - x^2 -1$. Its first derivative is $f'(x) = e^x - 2x$. Let's see for which $x \in [0, +\infty)$ $f'(x) > 0 \ $: differentiate $f$ once more and you obtain $f''(x) = e^x -2 \ > 0 \Leftrightarrow x > \ln(2)$. In other words $x = \ln(2) \ $ is a minimum for $f' \ $, which means that  $f'(x) \ge f'(\ln(2)) = 2 - 2\ln(2) > 0 \ $ because $\ln(2) < 1 \ $. Thus $f$ is a strictly monotone increasing function, which yields $f(x) > f(0) = 0 \  $ for all $x \in (0, +\infty) $.
A: One of many approaches (given you have tagged this as calculus):
For $0 \lt x \le 1$ compare $1+x^2$ with $1+x+x^2/2! + x^3/3! + \cdots$, noting $x \ge x^2$ in this interval
For $1 \lt x$ note $1+1^2 \lt e^1$ and compare $\frac{d}{dx} (1+x^2) = 2x$ with $\frac{d}{dx} e^x = e^x$, and if necessary note $2\times 1 \lt e^1$ and compare $\frac{d^2}{dx^2} (1+x^2) = 2$ with $\frac{d^2}{dx^2} e^x = e^x$.  
A: Using the series definition of $e^x$, we only have to prove that $1 + x + \dfrac{x^2}2 + \dfrac{x^3}6 \ge 1+x^2$ for all  $x\ge0$. This is equivalent to $x^2-3x+6 \ge 0$, which is true since the discriminant is negative.
A: By the AM-GM inequality
$$ \frac{2t}{1+t^2} < 1 $$
for any $t\in\mathbb{R}^+\setminus\{1\}$, hence by integrating both sides over $[0,x]$ we have
$$ \log(1+x^2) < x $$
as wanted.
A: Define the function $f(x)=e^x-x^2-1$
$f(0)=0$
Prove that the derivative is always positive then the function increases and if at zero is zero then after is positive.
(to prove the derivative is positive you can derive it and see that it has a minimum positive.
A: You need the elementary inequality for $x>0$
$$\frac{x}{x+1} \leq \log (1+x) \leq x $$
This yields
$$\frac{x^2}{x^2+1} \leq \log (1+x^2) \leq x^2 $$
which is what you want.
