Applications of the inequality $e^x\geqslant 1+x$ I am interested in proofs of famous theorems or inequalities which can be proved by the inequality
$$
e^x\geqslant 1+x.
$$  
For example, the divergence of harmonic series can be proved by assigning $x=\frac{1}{k}$.
$e^{\pi}>\pi^e$ can be proved by assigning $x=\frac{\pi}{e}-1$.  
The inequality of the arithmetic average of n-th degree can be proved by assigning $x=\frac{a_k}{(a_1*a_2*\cdots*a_n)^{\frac{1}{n}}} -1$.
The divergence of $\frac{e^t}{t^n}$ can be proved by assigning $x=\frac{t}{n+1}$.  

I want to know other possible proofs.
 A: Define $p_i$ as the i-th prime number. Taking $x=\frac{1}{p_i -1}$ and we have 
$$e^{\frac{1}{p_i-1}}≧1+\frac{1}{p_i -1}=1+\frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots.$$
The product from $i=1$ to $n$ is
$$e^{1+\sum_{i=2}^n \frac{1}{p_i-1}}≧\prod_{i=1}^n \left(1+ \frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots \right).$$
And here,
$$\prod_{i=1}^n \left(1+ \frac{1}{p_i} +\frac{1}{p_i ^2}+\cdots \right)≧\sum_{k=1}^{x(n)} \frac{1}{k},$$
$$\sum_{i=2}^n \frac{1}{p_i-1}≦ \sum_{i=1}^{n-1} \frac{1}{p_i}.$$  
So, we can say
$$e^{\sum_{i=1}^{n-1} \frac{1}{p_i}}≧ \sum_{k=1}^{x(n)} \frac{1}{k},$$
$$\Leftrightarrow\sum_{i=1}^{n-1} \frac{1}{p_i}≧\log \sum_{k=1}^{x(n)} \frac{1}{k}.$$ 
When $n\to\infty$, $x(n)\to\infty$ and $\lim_{n\to\infty} \log \sum_{k=1}^{x(n)} \frac{1}{k}=\infty$. So,
$$\sum_p \frac{1}{p}=\infty.$$ 
That is how we can prove the sum of prime numbers diverges.
A: *Taking $x= \log(t+1),~~t>0$ in that inequalty we have the following 
$$\color{blue}{\log(t+1)\le t, ~~\forall t>0}$$
*Also see here What is a nice way to prove that : $\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t}$
*Or taking $x=\log\left(\frac{1}{k^2}+1\right)$ we have 
$$\color{blue}{\log(\frac{1}{k^2}+1)\le \frac{1}{k^2}, ~~\forall k>0}$$
then the series 
$$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)\le\sum_{k=1}^{\infty}\frac{1}{k^2} =  \frac{\pi^2}{6}$$
converges and its sum is less that $\frac{\pi^2}{6}$

From this Is there a close form of: $\sum\limits_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)$ we have that, 
  $$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)= \log \frac{\sinh \pi}{\pi} $$

A: Interestingly the variant $\log(1+x) \leq x$ is widely used in Information Theory. For instance, you can use it to prove the log sum inequality:
For $a_k\geq 0$ and $b_k>0$,
$$\sum_{k=1}^n a_k\log \frac{a_k}{b_k} \geq  \sum_{k=1}^n a_k \log \frac{\sum_{l=1}^n a_l}{\sum_{j=1}^n b_j}$$
And as a corollary, the KL Divergence between two distributions $P$ and $Q$
$$D(P\|Q) \geq 0$$.
I'll update the list when I get more.
Update: The proof of the partial converse of Borel Cantelli lemma also uses this.
A: A standard use is that
$\prod (1+a_k)
\le \prod e^{a_k}
=  e^{\sum a_k}
$ 
so that,
if $\sum a_k$ converges then
$\prod (1+a_k)$
also converges.
