# Solution of $1+x<e^x$

Show that $$1+x for all $$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?

• Yes $100%$. {}{}{}{}{}{}{[{[{ – hamam_Abdallah Nov 15 '20 at 23:25
• alternative solution: $e^x=1+x+\dfrac {x^2}2+\dfrac {x^3}{3!}+\cdots=1+x+\text{positive terms}>1+x$, for $x>0$ – J. W. Tanner Nov 15 '20 at 23:31
• J.W. Tanner's solution is the simplest, although it depends what definition of $e$ you are using for this question... – Adam Rubinson Nov 15 '20 at 23:32

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 $$x\lt0$$.

• 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. – user926356 Nov 15 '20 at 23:36

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.

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