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Show that $1+x<e^x$ for all $x>0$

My attempt:

Consider the function $f(x)=e^x-x-1$ and fix $x>0$. By the MVT there exists a point $c\in(0,x)$ such that $\frac{f(x)-f(0)}{x-0}=f'(c)~~~(*)$.

Note that $f(0)=e^0-0-1=0$ and $f'(x)=e^x-1$. This implies that $f'(c)=e^c-1$.

The equality $(*)$ can be rewritten as $\frac{e^x-x-1-0}{x}=e^c-1$.

Since $c$ is a positive number, we have that $e^c>1$. So, the difference $e^c-1>0$.

This means that $\frac{e^x-x-1-0}{x}>0$. Then, $e^x>1+x$ for $x>0$.

Is this solution correct?

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    $\begingroup$ Yes $100%$. {}{}{}{}{}{}{[{[{ $\endgroup$ – hamam_Abdallah Nov 15 '20 at 23:25
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    $\begingroup$ alternative solution: $e^x=1+x+\dfrac {x^2}2+\dfrac {x^3}{3!}+\cdots=1+x+\text{positive terms}>1+x$, for $x>0$ $\endgroup$ – J. W. Tanner Nov 15 '20 at 23:31
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    $\begingroup$ J.W. Tanner's solution is the simplest, although it depends what definition of $e$ you are using for this question... $\endgroup$ – Adam Rubinson Nov 15 '20 at 23:32
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Your solution is fine, if a bit more complicated than necessary. If you are allowing derivatives, why not just note that $f'(x)=e^x-1\gt0$ for $x\gt0$, so $f$ is strictly increasing, hence $f(x)\gt f(0)=0$ for all $x\gt0$?

Just for fun, there is a non-calculus proof, provided you know that $(1+x/n)^n$ increases to $e^x$ as $n\to\infty$ (for $x\gt0$). Then the binomial expansion (with $n\gt1$) tells us

$$1+x=1+n\cdot{x\over n}\lt\left(1+{x\over n}\right)^n\lt e^x$$

for all $x\gt0$.

Added later: On further reflection, it occurs to me that your MVT approach actually works nicely as a proof that $e^x\gt1+x$ for all $x\not=0$, not just for positive $x$. That is, the Mean Value Theorem guarantees a $c$ between $0$ and $x$ such that $f(x)/x=f'(c)=e^c-1$. Now $x$ and $e^c-1$ have the same sign (e.g., if $x\lt c\lt0$, then $e^c\lt1$), so either way $f(x)$ must be positive. The other proofs here have to do something different to handle the case $x\lt0$.

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    $\begingroup$ Thank you for the alternative solution. I'm familiar with the last statement, but I never thought of using it to prove the desired inequality. $\endgroup$ – user926356 Nov 15 '20 at 23:36
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Consider the function $f(x) = e^{x} - x - 1$. Consider $f'(x) = e^{x} - 1$, and note that this is positive for all $x > 0$ because $f'(0) = 0$ and $e^{x}$ is an increasing function. Thus, $f(x)$ is strictly increasing for all $x>0$. However, since $f(0) = 0$, this means that $f(x)>0$ for all $x > 0$, and thus:

$$\boxed{e^{x} > x + 1\text{ for }x>0}$$

We can extend this to $e^{x} > x + 1$ for $x < 0$ using a similar argument.

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Others have already commented on your solution.

Alternatively, if $x>0$, then $e^x=1+x+\dfrac{x^2}2+\dfrac{x^3}{3!}+\cdots=1+x+\text{ positive terms } > 1+x$.

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