Prove $\exp(x) \geq 1+x \forall x \in \mathbb{R}$

Question

I've managed to prove the statement for $x \geq 0 $ and $x \leq -1$ but I can't manage to construct a proof for $ -1 < x < 0 $

My lecture done it by proving $ \exp(x) - (1+x) = \displaystyle \sum_{n=2}^\infty \dfrac{x^n}{n!} $ and showing that is greater than or equal to 0. For $ -1 < x < 0 $ he set $a=-x$ and then considered for $0 < a < 1$ but I really don't understand his steps from there onwards.

Sorry I should have added this from the start - Yes, we defined $\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$ We also have proved that $\exp(x+y) = \exp(x)\exp(y) $ - I'd rather not use differentiation as when we covered this we had not yet touched calculus.

Answer 1

To proceed from the series, write the summation in the following way: $$\exp (x)-(1+x)=\sum_{k=1}^{\infty} \left(\frac{x^{2k}}{2k!} + \frac{x^{2k+1}}{(2k+1)!} \right)$$ notice the summand is positive for $-1<x<0$, since $|x|^{2k} >|x|^{2k+1},\ x^{2k}>0>x^{2k+1},\ 1/(2k)!>1/(2k+1)!$

Answer 2

Hint: consider the function $$ f(x)=e^x-1-x $$ and find its minimum.

Answer 3

Define

$$f(x):=e^x-x-1\implies f'(x)=e^x-1=0\iff x=0$$

But

$$f''(x)=e^x\implies f''(0)=e^0=1>1\implies\;\text{at $\;x=0\;$ we have a minimum, and thus}$$

$$e^x-x-1=f(x)\ge f(0)=1-0-1=0\implies e^x\ge 1+ x$$

Answer 4

The function $f\colon x\mapsto e^x$ is convex hence its curve is above the tangent line at the point $x=0$ with equation $$y=f'(0)x+f(0)=x+1$$ so we have the desired inequality.

Answer 5

For all $x>-1$ $$\log(1+x) = \int_1^{1+x}\frac{dt}{t} \leq x.$$ Now take $x = e^x-1$.