AM-GM inequality

Let $a_1,\cdots,a_n$ be positive real numbers. Then $$\sqrt[n]{a_1\cdots a_n}\leq \dfrac{a_1+\cdots+a_n}{n},\qquad (*)$$ with equality iff all $a_k\:(1\leq k\leq n)$ are equal.

Proof by standard induction
Let $H(n)$ stand for the hypothesis that inequality $(*)$ is valid for $n$. For $n=2$, we have $$\sqrt{a_1a_2}\leq \dfrac{a_1+a_2}{2}.\qquad (**)$$ This follows from $(\sqrt{a_1}-\sqrt{a_2})^2\geq 0$. Moreover, equality in $(**)$ holds iff $a_1=a_2$. Now we will show that $H(2)$ and $H(n)$ together imply $H(n+1)$. Let $$A_{n+1}=\dfrac{a_1+\cdots+a_n+a_{n+1}}{n+1}.$$ Then $$\begin{align}A_{n+1}&=\dfrac{1}{2n}[(n+1)A_{n+1}+(n-1)A_{n+1}]\\&=\dfrac{1}{2n}[(a_1+\cdots+a_n)+(a_{n+1}+A_{n+1}+\cdots+A_{n+1})]\\&\geq \dfrac{1}{2n}(n \sqrt[n]{a_1\cdots a_n}+n\sqrt[n]{a_{n+1}A_{n+1}^{n-1}})\\&\geq \sqrt[2n]{a_1\cdots a_n a_{n+1} A_{n+1}^{n-1}},\end{align}$$ where the first inequality follows from $H(n)$, while the second inequality follows from $H(2)$. Now, $H(n+1)$ follows from $$A_{n+1}^{2n}\geq a_1\cdots a_{n+1}A_{n+1}^{n-1}.$$ Equality holds iff $a_1=\cdots=a_n$ and $a_{n+1}=A_{n+1}$, or $a_1=\cdots=a_{n+1}$. $\Box$

Proof by analysis
The inequality $$\inf x +\inf y \leq \inf (x+y)$$ can be generalised to $$\sum\limits_{i=1}^n \inf f_i(x) \leq \inf \sum\limits_{i=1}^n f_i(x).$$ [Below we used $\sum$ to mean $\sum\limits_{i=1}^n$.]
Taking log on both sides of $(*)$, $$\sum \ln a_i\leq n \ln(\dfrac{\sum a_i}{n}).$$
Adding $n$ on both sides, $$\sum (1+\ln a_i)\leq n+n \ln(\dfrac{\sum a_i}{n}).$$
Claim $$f_i(x)=a_ix-\ln x\;\text{and}\;\min f_i(x)=1+\ln a_i.$$ Then $f_i'(x)=a_i-\frac{1}{x}$ and $f_i''(x)=\dfrac{1}{x^2}>0.$ It follows that $\min f_i(x)=1+\ln a_i$.
Now suppose $$f(x)=\sum f_i(x)=(\sum a_i)x-n\ln x.$$ Then $f'(x)=\sum a_i-\frac{n}{x} \Rightarrow x=\dfrac{n}{\sum a_i}$ and $f''(x)=\dfrac{n}{x^2}>0$. It follows that $\min f(x)=n+n\ln \dfrac{\sum a_i}{n}$.
Therefore $n+\ln a_1+\cdots+\ln a_n \leq n+n\ln \dfrac{\sum a_i}{n}\Rightarrow \dfrac{1}{n}\ln(a_1\cdots a_n)\leq \ln\dfrac{\sum a_i}{n}.\;\Box$

An application
Prove that $\{(1+\frac{1}{n})^n\}$ is monotonic increasing and bounded.
Proof: By the AM-GM inequality, we have $$\sqrt[n+1]{(1+\frac{1}{n})^n\cdot 1}<\dfrac{1}{n+1}[n(1+\frac{1}{n})+1]=1+\dfrac{1}{n+1}.$$ This is equivalent to $$(1+\dfrac{1}{n})^n<(1+\dfrac{1}{n+1})^{n+1},$$ which shows that $\{(1+\frac{1}{n})^n\}$ is monotonic increasing as required. To find an upper bound of the sequence, for $n>5$, using the AM-GM inequality once more, we have $$\sqrt[n+1]{(\dfrac{5}{6})^6\cdot 1^{n-5}}<\dfrac{1}{n+1}[6\cdot \dfrac{5}{6}+(n-5)\cdot 1]=\dfrac{n}{n+1},$$ and so $(\dfrac{5}{6})^6<(\dfrac{n}{n+1})^{n+1}$, or equivalently, $(1+\dfrac{1}{n})^{n+1}<(\dfrac{6}{5})^6$. Therefore, $$(1+\dfrac{1}{n})^n<(1+\dfrac{1}{n})^{n+1}<(\dfrac{6}{5})^6<3,\quad (\text{for}\; n>5).$$ Since $(1+\dfrac{1}{n})^n<3$, we conclude that $\{(1+\frac{1}{n})^n\}$ is bounded. $\Box$

More applications

Reference:
Excursions in Classical Analysis: Pathways to Advanced Problem Solving and Undergraduate Research