Applications of the inequality $e^x\geqslant 1+x$

Question

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 $x=\frac{t}{n+1}$.

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I want to know other possible proofs.

Answer 1

*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, $$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)= \log \frac{\sinh \pi}{\pi} $$

Answer 2

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.

Answer 3

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.

Answer 4

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.