Going trough my math textbook, I stumbled upon a proof of inequality of number means i.e. $$ H_n \le G_n \le A_n \le Q_n $$ where $H_n, G_n , A_n , Q_n$ are harmonic, geometric, arithmetic and quadratic means of $n$ real numbers ($n\in \mathbb{N})$. The textbook states:
This is one of the most important inequalities in mathematics. There are over 40 different known proofs of this inequality.
I would like to know more about such great importance of this inequality and it's appliances in problem solving and other branches of science. Also, I would be grateful if someone were to provide few hyperlinks with one or more different proofs to this inequality.
The question a little bit too board. I don't know what is the $40$ different proof, but now I will give you two classical proofs: one for $H_n \leq G_n$, and one for $G_n \leq A_n$. After that I will give a general approach, which give us proofs for whole family of this types of inequalities. At last I will write a little bit about applications.
Classic results on inequalities of means.
First of all we define the harmonic ($H_n$), geometric ($G_n$), arithmetic ($A_n$) and quadratic ($Q_n$) means as the following. $$ \begin{align} H_n(x_1,\dots,x_n) & := \frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}};\\ G_n(x_1,\dots,x_n) & := \sqrt[n]{x_1\cdots x_n};\\ A_n(x_1,\dots,x_n) & := (x_1+\cdots+x_n)/n;\\ Q_n(x_1,\dots,x_n) & := \sqrt{ \frac{1}{n} \left( x_1^2 + \cdots + x_n^2 \right) }. \end{align} $$
Proof. If there exists and $x_j$ which is zero, then the statement is trivial. The case $n=1$ is also trivial. With mathematical induction, if the statement is true for $n-1$, then because $\ln$ function is concave, if we introduce the notation $x:=\sqrt[n-1]{x_1\cdots x_{n-1}}$, then $$\frac{n-1}{n}\ln(x) + \frac{1}{n}\ln(x_n) \leq \ln\left(\frac{n-1}{n}x+\frac{1}{n}x_n\right).$$ From here $$\sqrt[n]{x^{n-1}x_n} \leq \frac{1}{n}x_n + \frac{n-1}{n}x \leq \frac{1}{n}x_n + \frac{n-1}{n} \frac{x_1+\cdots+x_{n-1}}{n-1},$$ and the equity holds exactly when $x_1 = \cdots = x_n$. $\square$
Proof. Let $x_1,\dots,x_n$ be positive real numbers. Apply Theorem $1$ for the also positive real numbers $1/x_1,\dots,1/x_n$. $\square$
General results on inequalities of means.
Now, we will show theorems from which the result
naturally comes.
As mentioned earlier, we wouldn't prove all the applied results, but the following theorem is the most important behind all the following results.
Theorem $3$: Jensen's inequality. Let $p_1, \dots p_n$ be nonnegative real numbers for that $p_1 + \cdots p_n = 1$. Let $I \subset \mathbb{R}$ be an interval and $f : I \rightarrow \mathbb{R}$ a convex function. In this case the Jensen's inequality holds: $$f\left(\textstyle\sum_{i=1}^n p_i q_i \right) \leq \sum_{i=1}^n p_i f(q_i),$$ where $q_1,\cdots,q_n \in I$. If $f$ is strictly convex, then the equity holds if and only if $q_1 = \cdots = q_n$.
Proof. We will use mathematical induction. If $n=1$, then the statement is trivial. The case of $n=2$ is the definition of convexity. With induction let $\sum_{i=1}^{n+1} p_i = 1$. Then $$ \begin{split} f\left(\textstyle\sum_{i=1}^{n+1} p_i q_i \right) = f\left(\textstyle\sum_{i=1}^{n} p_i q_i + p_{n+1} q_{n+1} \right) \\ \text{(inductive hypothesis)} \quad \quad \quad \quad \!\!\! \leq \left(\sum_{k=1}^n p_k \right) \cdot f\left( \frac{\sum_{i=1}^n p_i q_i}{\sum_{j=1}^n p_j} \right) + p_{n+1} f(q_{n+1}) \\ \text{(def. of convexity)} \quad \leq \left(\sum_{k=1}^n p_k \right) \cdot \left( \sum_{i=i}^n \left( \frac{p_i}{\sum_{j=1}^n p_j} f(q_i) \right) \right) + p_{n+1}f(q_{n+1}) \\ = \sum_{i=1}^{n+1} p_i f(q_i), \end{split} $$ furthermore, if $f$ is strictly convex, then the equity holds exactly when $q_1 = q_2 = \cdots = q_n$, and $q_{n+1} = \sum_{i=1}^n p_i q_i / \sum_{j=1}^n p_j = q_n$. $\square$
Remark. If $f$ is concave, then the statement is true for the reverse inequality. The proof of this remark is analogous with the proof of Jensen's inequality.
The definition of abstract means was introduced be Cauchy.
Definition: Mean. Let $I \subseteq \mathbb{R}_+$ be an open interval, and $n \in \mathbb{N}_+$. The $\mathfrak{M} : I^n \rightarrow \mathbb{R}_+$ function is mean, if
There are a lot of other different mean definition. Some author define symmetry in a different way, and there are also theorems to charaterise class of functions which satisfy given properties.
In 1930 independently B. Finetti, A. Kolmogorov, and M. Nagumo defined and studied the following concept.
Definition: Quasi-arithmetic mean. Let $I \subseteq \mathbb{R}_+$ be an open interval and $n \in \mathbb{N}_+$. Furthermore let $\varphi : I \rightarrow \mathbb{R}$ be a continous and strictly monotonic function. Then the $\mathfrak{M}_\varphi : I^n \rightarrow \mathbb{R}$ quasi-arithmetic mean is defined as $$ \mathfrak{M}_\varphi (x_1, \dots, x_n) := \varphi^{-1}\left(\frac{1}{n} \sum_{i=1}^n \varphi(x_i) \right). $$
$H_n,G_n,A_n,Q_n$ are special cases:
For $\varphi,\psi : I \rightarrow \mathbb{R}$ continous and strictly monotonic functions, for $a\neq 0$ and $b$ real numbers for the linear transformation $\psi = a\varphi + b$ there is $\mathfrak{M}_\varphi = \mathfrak{M}_\psi$.
The following two results are from
and from
We transform the results from the paper from continous into a discrete statement.
Proof. Parametrize the Jensen's inequality with $f:=\psi \circ \varphi^{-1}$ and $I$ is and appropriate interval to the means, and $q_i := \psi(x_i)$ for all $i=1,\dots,n$. Then $$\psi \circ \varphi^{-1}\left( \sum_{i=1}^n p_i \cdot \varphi(x_i) \right) \leq \sum_{i=1}^n p_i \cdot \psi \circ \varphi^{-1}(\varphi(x_i)).$$ Because $\psi^{-1}$ is monotonically increasing, we could apply is to both sides and this way we get $$\varphi^{-1}\left( \sum_{i=1}^n p_i \cdot \varphi(x_i) \right) \leq \psi^{-1}\left( \sum_{i=1}^n p_i \cdot \psi(x_i) \right).$$ From here because of the definition of quasi-arithmetic means, if $p_i = 1/n$ for all $i=1,\dots,n$ we get $\mathfrak{M}_{\varphi} \leq \mathfrak{M}_\psi$. For the case $\psi \circ \varphi^{-1}$ we could apply Jensen's inequality in the concave case. And the rest of the statment is analagous. $\square$
Deriving the inequality chain from Theorem $4$. For $H_n \leq G_n$ let $I:=\mathbb{R}_+$, $\psi(x) := \ln(x)$ and $\varphi(x):=1/x$. In this case $\psi^{-1}(x)=\exp(x)$ and $\varphi^{-1}(x)=1/x$ on $\mathbb{R}_+$. Because $\psi \circ \varphi^{-1}\left(x\right) = \ln(1/x)$ function is convex and $\psi^{-1}=\exp(x)$ is monotonically increasing we get that $H_n \leq G_n$. With appropriate parameters we could derive the other inequalities. With $I:=\mathbb{R}_+$, $\psi(x) := x$ and $\varphi(x):=\ln(x)$ we get the result $G_n \leq A_n$ and with $I:=\mathbb{R}_+$, $\psi(x) := x^2$ and $\varphi(x):=x$ we get the result $A_n \leq Q_n$.
I let the proof and the appropriate deriving to the reader. $\square$
Applications.
You could find applications in economics, for example, in financial analysis. Using means are also important in decision theory.