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

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  • $\begingroup$ As $e^x\ge x+1$ can be proved without using differentiation and logarithm, that $e^x-x>0$ should be an easy problem. $\endgroup$
    – user1551
    Jan 8, 2013 at 11:46
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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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  • $\begingroup$ The same argument show you that $e^x \geq x + 1$, and therefore $e^x > x$. $\endgroup$ Jan 8, 2013 at 12:50
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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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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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You can also check the taylor series for $e^x$ when $x>0$

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  • $\begingroup$ That obviously works for $x \geq 0$, but a tiny bit more work is required for $x < 0$. $\endgroup$ Jan 8, 2013 at 14:49
  • $\begingroup$ @Andrew Uzzell The inequality is obviously true when $x<0$ $\endgroup$
    – Amr
    Jan 8, 2013 at 23:06
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$$e^x - x >0\iff e^x >x$$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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  • $\begingroup$ "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. $\endgroup$
    – Adam
    Jan 8, 2013 at 11:43
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For $x>0$ we have $e^t>1$ for $0<t<x$

Hence, $$x=\int_0^x1dt \color{red}{\le} \int_0^xe^tdt =e^x-1 \implies 1+x\le e^x\implies0< x-e^x$$ For $x<0$ we have $e^{t} <1$ for $x <t<0$

$$-x=\int^0_x1dt \color{red}{\ge} \int^0_xe^tdt =1-e^x \implies 1+x\le e^x\implies0< x-e^x$$

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