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Prove that the following inequality is true for all real numbers $$e^x-x>0.$$

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Michael Albanese Jan 8 '13 at 11:12
As $e^x\ge x+1$ can be proved without using differentiation and logarithm, that $e^x-x>0$ should be an easy problem. – user1551 Jan 8 '13 at 11:46
I thank you for those remarks and apologize to all (because, beginning in the English language) I found the site sobering and effective and you look for different solutions to tack questions I will try to be an active member Please accept my apologies Thank you for all – Ahmed Jan 8 '13 at 18:39

Let $f(x)=e^x-x$. Then obviously $f(x)>0$ for $x< 0$. Also we have that $$ f'(x)=e^x-1\geq 0,\quad x\geq 0 $$ so $f$ is increasing on $[0,\infty)$ and because $f(0)=1$ we see that $f(x)\geq 1>0$ for $x> 0$.

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First if $e^x = x$ then $x > 0$, but $(e^x)' > (x)'$ if $x > 0$ and so $e^x$ increase faster than $x$. Finally observe that $e^0 > 0$.

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The same argument show you that $e^x \geq x + 1$, and therefore $e^x > x$. – Diego Silvera Jan 8 '13 at 12:50

The inequality is obviousy true for neqgative values of $x$.

$e^0-0=1>0$ and $\frac{d(e^x-x)}{dx}=e^x-1\geq0$. Thus, $e^x-x$ is increasing on $[0,\infty)$. Thus, $\forall x>0[e^x-x>e^0-0=1>0]$

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Let $f(x)=e^x-x$. You have that at $x=0$, $f(0)=e^0-0=1$. You also have that $f'(x)=e^x-1$. Since $e^x$ is strictly increasing and since $e^0=1$ you have that for all $x>0$, $f'(x)>0$. Since $f(0)=1>0$ and since for $x>0$ the function is increasing, it follows that $f(x)>0$ for all $x>0$.

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You can also check the taylor series for $e^x$ when $x>0$

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That obviously works for $x \geq 0$, but a tiny bit more work is required for $x < 0$. – Andrew Uzzell Jan 8 '13 at 14:49
@Andrew Uzzell The inequality is obviously true when $x<0$ – Amr Jan 8 '13 at 23:06

$$e^x - x >0\iff e^x >x$$It comes down to proving that $e^x$ is always increasing, and faster than $x$. Recall, the definition of an increasing function is a function which always has a positive slope. This means that we should prove that $e^x > 0 $ for all $x$.

Now, $\dfrac{d}{dx} e^x = e^x$ which means that the slope of $e^x$ is always $e^x$ and we know that $e^x$ will always be positive. Thus $f(x) = e^x$ will always have a positive slope. We have proved that $e^x$ is always increasing, so now consider these cases:

  • $x$ is positive in $e^x$: We know that $e^x$ is always increasing faster than $x$, so $e^x > x$ for positive $x$. Simplifying, $e^x -x >0$.

  • $x$ is $0$ in $e^x$: Proof by cases (only one case). $e^0 = 1 > 0$ or $e^x > x \iff e^x - x > 0$ in this case.

  • $x$ is negative in $e^x$: Let $x = -k$. Then $e^{-k} = \dfrac{1}{e^k}$. We know that $k$ is positive so $\dfrac{1}{e^k}$ will never be negative since $e^k$ is positive for positive $k$. So $e^{-k}$ is always positive and $-k $ (which we know is $x$) is negative. So $e^{-k} > -k \iff e^x > x \iff e^x - x > 0$.

In all three cases, we get $e^x - x > 0$ which means that the statement $e^x -x >0$ is always true.

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"We know that $e^x$ is always increasing, so $e^x>x$ for positive $x$". $g(x)=\frac{x}{2}+1$ is also increasing (and also $g(0)=1>0$) but it's not true that $\frac{x}{2}+1 > x$ for all $x$. What we need to know is that $e^x$ is increasing faster than $x$, not only that it's increasing. – Adam Jan 8 '13 at 11:43

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