Showing that $e^x > 1+x$

Problem.

Show that $$e^x > 1 + x , \ \ x \neq 0$$

My attempt.

Using Mean Value Theorem: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$

$$\Rightarrow e^c = \frac{e^x - 1}{x}$$ $$\Rightarrow xe^c = e^x -1$$ $$\Rightarrow x+1 = \frac{e^x}{e^c}$$

I'm not feeling very comfortable with these calculations...Am I on the right track? Can I even choose $e^o$ in my initial calculation as it says that $x \neq 0 $ ?

Solution 1:

A geometric way to prove the inequality is to use the convexity of the function $f=\exp$ (since its second derivative is $e^x\ge0$) then the line tangent at $x=0$ with equation: $$y=f'(0)x+f(0)=x+1$$ is below the curve of the function hence $$e^x> x+1,\quad x\ne0$$

Solution 2:

The Mean Value Theorem is a valid approach. Here is another approach; hopefully, it will make you more comfortable to have two.

The integer version of Bernoulli's Inequality, proven at the end of this answer, is sufficient to prove that, for $n_0\gt\max(-x,1)$ and $x\ne0$, we have $$ 1+x\lt\left(1+\frac x{n_0}\right)^{n_0}\le\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x $$

Solution 3:

for $x>0$, $e^x>1$ so $$\int_0^b e^xdx>\int_0^b dx=b$$ $$e^b-1>b$$ $$e^b>b+1$$ for $x<0$, $e^x<1$ so $$\int_a^0 e^xdx<\int_a^0dx,a<0$$ $$1-e^a<-a$$ $$1+a<e^a$$ so for any non zero number $c, e^c>c+1$ seeing as $e^0=1$