Prove that $e^x>x+1 \forall x\ne 0$

Question

I need to prove that $e^x>x+1 \forall x\ne 0$.

Any ideas of hints about how to begin? I don't have any idea except the graphical way.

Answer 1

Hint

$$\frac{d}{dx} (e^x-x-1)=e^x-1.$$

Then try to study the variations of $x\mapsto e^x-x-1$.

Answer 2

By the Bernoulli inequality, $$1+x\leq \left(1+\frac{x}{n}\right)^n$$ and $$\left(1+\frac{x}{n}\right)^n\rightarrow e^x.$$

An interesting way to prove strictness is also by Bernoulli, If $$a^x=x+1$$ then $$1+x(a-1)\leq a^x=1+x$$ and thus $a\leq 2$. $e$ however is $>2$.

In fact $\left(1+\frac{x}{n}\right)^n$ is increasing for all $x$. And the first term in the sequence is $1+x$. The strict inequality for all $x$ then follows from $1+x<\left(1+\frac{x}{2}\right)^2$ for all $x\neq 0$.

Answer 3

Let

$$g(t)=\ln(t+1)-t$$

for $t\geq0$.

$$\forall t>0\;\; g'(t)=-\frac{t}{1+t}<0$$

$$\implies \forall t>0\;\; \;\ln(t+1)-t<g(0)=0$$

$$\implies \forall t>0\;\;\; t+1<e^t$$

Answer 4

$$\frac{e^x}{x+1}>1$$ $$1+\frac{\frac{x^2}{2!}+\frac{x^3}{3!}+....}{x+1}>1$$