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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 $\frac{x^2}{1+x^2}\leq \ln (1+x^2)\leq x^2$?2012-01-25
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    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?2012-01-25
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    Think about the derivatives.2012-01-25
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    What is your definition of $e^x$?2012-01-25
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    @lhf $ln e = 1$2012-01-25
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    Ok, $\ln e =1 $ defines $e$. How do you define $\ln(x)$ and $a^x$ then? I ask for definitions because if you had defined $e^x$ to be a series then the answer to your question would be immediate.2012-01-25
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    @ChrisTaylor and prove that $e^x$ derivative is always larger? I tried that for a bit but didn't manage very well maybe I missed somthing2012-01-25
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    This is a 'classic' case of using calculus to prove some inequality. What do want to differentiate?2012-01-25
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    Is it possible to use induction on unit intervals (0,1], (1,2],...? If it can be done, is there a good example where it works nicely?2012-01-25
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    http://mathoverflow.net/questions/38238/a-principle-of-mathematical-induction-for-partially-ordered-sets-with-infima2012-01-25

5 Answers 5

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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 lot2012-01-25
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    I'm glad it helped :)2012-01-25
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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 :) Thanks2012-01-25
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    actually I don't think i understand the part for $0 can you elaborate on that?2012-01-25
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    For $0 \lt x \le 1$ you have $x \ge x^2$ so $$1+x^2 \le 1+x \lt 1+x+x^2/2! + x^3/3! + \cdots$$2012-01-25
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    Ok cool, but how does that prove my inequality? Sorry for the probably obvious questions...2012-01-25
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    The left hand side is $1+x^2$ and the right hand side is [$e^x$](http://en.wikipedia.org/wiki/Exponential_function#Formal_definition)2012-01-25
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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.