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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.

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    $\begingroup$ Note $A \ge G \iff G \ge H$ by the transform $x_i \mapsto \dfrac1{x_i}$. Similarly $A \ge G \implies Q \ge A$ as $$x_i^2 + x_j^2 \ge 2x_i x_j \implies n \sum x_i^2 \ge (\sum x_i)^2 \implies Q \ge A$$ so any proof of AM-GM generates the whole chain. There are many proofs for AM-GM on the net, check math.stackexchange.com/questions/691807/… for a few. $\endgroup$
    – Macavity
    Dec 8 '14 at 5:52
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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. $$ \begin{align} H_n(x_1,\dots,x_n) & := \frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}};\\ G_n(x_1,\dots,x_n) & := \sqrt[n]{x_1\cdots x_n};\\ A_n(x_1,\dots,x_n) & := (x_1+\cdots+x_n)/n;\\ Q_n(x_1,\dots,x_n) & := \sqrt{ \frac{1}{n} \left( x_1^2 + \cdots + x_n^2 \right) }. \end{align} $$

Theorem $1$: Inequality of arithmetic and geometric means. If $n \in \mathbb{N}_+$, $x_1,\dots,x_n \geq 0$, then $\sqrt[n]{x_1\cdots x_n} \leq (x_1+\cdots+x_n)/n$, so $G_n \leq A_n$, furthermore the equity holds if and only if $x_1 = \cdots = x_n$.

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 $x_1 = \cdots = x_n$. $\square$

Theorem $2$: Inequality of geometric and harmonic means. If $n \in \mathbb{N}_+$, $x_1,\dots,x_n \geq 0$, then $\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} \leq \sqrt[n]{x_1\cdots x_n}$, so $H_n \leq G_n$, furthermore the equity holds if and only if $x_1 = \cdots = x_n$.

Proof. Let $x_1,\dots,n_n$ be positive real numbers. Apply Theorem $1$ for the also positive real $1/x_1,\dots,1/x_n$ numbers. $\square$

General results on inequalities of means.

Now we will show theorems from which the result

$$H_n \leq G_n \leq A_n \leq Q_n$$

naturally comes.

As earlier we wouldn't proof all the applied results, but the following theorem is the most importent 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 $f$ 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 inequalty. 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

  • $\min(x_1, \dots, x_n) \leq \mathfrak{M}(x_1, \dots, x_n) \leq \max(x_1, \dots, x_n)$ $\quad$ (Cauchy's axiom of means);
  • $\mathfrak{M}(x_1, \dots, x_n) = \mathfrak{M}(x_{\sigma(1)}, \dots, x_{\sigma(n)})$, where $\sigma : \mathbb{N} \rightarrow \mathbb{N}$ is a permutation (symmetry);
  • $\mathfrak{M}(x, \dots, x) = x$ where the arity of $\mathfrak{M}$ is $n$ $\quad$ (reflexivity).

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 $\varphi_H(x) = 1/x$: $\quad\mathfrak{M}_{\varphi_H} = H_n$;
  • for $\varphi_G(x) = \ln(x)$:$\quad\mathfrak{M}_{\varphi_G} = G_n$;
  • for $\varphi_A(x) = x$: $\quad\mathfrak{M}_{\varphi_A} = A_n =: \mathfrak{M}_{\operatorname{id}}$;
  • for $\varphi_Q(x) = x^2$: $\quad\mathfrak{M}_{\varphi_Q} = Q_n$.

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

Micic, J. $-$ Pecaric, J. $-$ Seo, Y.: Order among quasi-arithmetic means of positive operators. Math. Reports, submitted for publication.

and from

Micic, J. $-$ Pavic, Z. $-$ Pecaric, J.: Jensen’s inequality for operators without operator convexity. Linear Algebra and its Applications, 434, pp. 1228$-$1237, 2011.

We transform the results from the paper from continous into a discrete statement.

Theorem $4$. Let $I \subseteq \mathbb{R}_+$ be an open interval, $\varphi,\psi: I \rightarrow \mathbb{R}$ continous and strictly monotonic functions. We define for them the appropriate $\mathfrak{M}_\varphi$ and $\mathfrak{M}_\psi$ quasi-arithmetic means. Then is one of the following is true:

  • $\psi \circ \varphi^{-1}$ function convex and $\psi^{-1}$ monotonically increasing, or
  • $\psi \circ \varphi^{-1}$ function concave and $\psi^{-1}$ monotonically decreasing, then $$\mathfrak{M}_\varphi \leq \mathfrak{M}_\psi.$$

And if one of the following is true:

  • $\psi \circ \varphi^{-1}$ function concave and $\psi^{-1}$ monotonically increasing, or
  • $\psi \circ \varphi^{-1}$ function convex and $\psi^{-1}$ monotonically decreasing, then $$\mathfrak{M}_\varphi \geq \mathfrak{M}_\psi.$$

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 $A_n \leq Q_n$.

Theorem $5$. Let $I \subseteq \mathbb{R}_+$ be an open interval, $\varphi,\psi: I \rightarrow \mathbb{R}$ continous and strictly monotonic functions. We define for them the appropriate $\mathfrak{M}_\varphi$ and $\mathfrak{M}_\psi$ quasi-arithmetic means. Then

  • if $\varphi,\psi$ strictly monotonic increasing functions, furthermore

    • if $\varphi$ concave and $\psi$ convex, then $\mathfrak{M}_\varphi \leq \mathfrak{M}_{\operatorname{id}} \leq \mathfrak{M}_\psi$, and
    • if $\varphi$ convex and $\psi$ concave then $\mathfrak{M}_\varphi \geq \mathfrak{M}_{\operatorname{id}} \geq \mathfrak{M}_\psi$,
  • if $\varphi,\psi$ strictly monotonic decreasing functions, furthermore

    • if $\varphi$ concave and $\psi$ convex, then $\mathfrak{M}_\varphi \geq \mathfrak{M}_{\operatorname{id}} \geq \mathfrak{M}_\psi$, and
    • if $\varphi$ convex and $\psi$ concave then $\mathfrak{M}_\varphi \leq \mathfrak{M}_{\operatorname{id}} \leq \mathfrak{M}_\psi$.

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.

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