# 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 :)

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 How do you deduce $x\leq\frac{x^2}{1+x^2}$ from $\frac{x^2}{1+x^2}\leq \ln (1+x^2)\leq x^2$? – Davide Giraudo Jan 25 '12 at 9:59 Ah my bad I made a mistake and confused $x>\ln (1+{{x}^{2}})$ with $x<\ln (1+{{x}^{2}})$ - But that still leaves me with $x\le x^2$ no? – Jason Jan 25 '12 at 10:02 Think about the derivatives. – Chris Taylor Jan 25 '12 at 10:08 What is your definition of $e^x$? – lhf Jan 25 '12 at 10:12 @lhf $ln e = 1$ – Jason Jan 25 '12 at 10:17

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 $x \in (0, +\infty)$.

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 I Like this method :) thanks a lot – Jason Jan 25 '12 at 13:58 I'm glad it helped :) – Emilio Ferrucci Jan 25 '12 at 14:03

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 $\frac{d^2}{dx^2} e^x = e^x$.

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I think I understood you, I will try to write this down and see if I got it completely :) Thanks – Jason Jan 25 '12 at 10:32
actually I don't think i understand the part for $0<x<1$ can you elaborate on that? – Jason Jan 25 '12 at 10:40
For $0 \lt x \le 1$ you have $x \ge x^2$ so $$1+x^2 \le 1+x \lt 1+x+x^2/2! + x^3/3! + \cdots$$ – Henry Jan 25 '12 at 11:32
Ok cool, but how does that prove my inequality? Sorry for the probably obvious questions... – Jason Jan 25 '12 at 11:49
The left hand side is $1+x^2$ and the right hand side is $e^x$ – Henry Jan 25 '12 at 12:04

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.

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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.

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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.

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