A short and interesting paper try to derive a series of inequalities from a basic theorem (the master theorem as dubbed in the paper):

A strictly concave function $$g:\mathbb{R}_+\to\mathbb{R}$$ and another function $$f:\mathbb{R}_+^2\to\mathbb{R}$$ defined as $$f(x,y)=y\cdot g(x/y)$$. For all positive real numbers $$x_1,\cdots,x_n$$ and $$y_1,\cdots,y_n$$ satisfy the inequality

$\sum_{i=1}^n f(x_i,y_i) \le f(\sum_{i=1}^n x_i, \sum_{i=1}^n y_i)$

with equality holds iff the two sequences are proportional, i.e., $$x_i/y_i=t$$ for some constant $$t$$. If the function $$g()$$ is concave but not strictly, the condition for equality relaxed. This theorem can be generalized into higher dimension for $$f(x_1,x_2,\cdots,x_d) = x_d g(x_1/x_d,x_2/x_d,\cdots,x_{d-1}/x_d)$$.

Concave function is satisfies the following inequalities on affine combination:

$\lambda g(x) + (1-\lambda)g(y) \le g(\lambda x+(1-\lambda)y)$

where $$\lambda\in[0,1]$$. In higher dimension, the function arguments $$x,y$$ can be vectors in $$\mathbb{R}_+^d$$. A twice-differentiable concave function has its second derivative nonpositive (or in high dimension, the Hessian matrix is negative semidefinite). Convex functions reverse the inequality sign above.

This paper concerns about concave function but it also mentioned two results of convex functions:

#### AM > GM

$\frac{1}{n}\sum_{i=1}^n a_i \ge \left(\prod_{i=1}^n a_i\right)^{1/n}$

#### Jensen’s inequality

$\frac{1}{n}\sum_{i=1}^n f(x_i) \ge f\left(\frac{1}{n}\sum_{i=1}^n x_i\right)^{1/n}$

for convex function $$f:\mathbb{R}\to\mathbb{R}$$

Below is how the different inequalities can be derived from the master theorem:

#### Cauchy-Schwarz inequality

$\sum_{i=1}^n a_i b_i \le \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{i=1}^n b_i^2}$

Normally this is proved by expansion of $$\sum(a_ib_j - a_jb_i)^2 \ge 0$$ but we can deduce this immediately from setting $$g(x)=\sqrt{x}$$ in the theorem, with $$f(x,y)=\sqrt{x}\sqrt{y}$$.

#### Hölder’s inequality

$\sum_{i=1}^n a_ib_i \le \left(\sum_{i=1}^n a_i^p\right)^{1/p} \left(\sum_{i=1}^n b_i^q\right)^{1/q}$

on the condition that $$p,q>1$$ and $$1/p+1/q=1$$. Which $$p=q=2$$ reduces Hölder’s inequality into Cauchy-Schwarz inequality.

Paper mentioned that the Hölder’s inequality can be proven by using Young’s inequality, namely, $$xy \le x^p/p + y^q/q$$ but we can also deduce this immediately by using $$g(x)=x^{1/p}$$ with $$f(x,y)=x^{1/p}y^{1/q}$$.

#### Minkowski inequality

$\left(\sum_{i=1}^n (a_i+b_i)^p\right)^{1/p} \le \left(\sum_{i=1}^n a_i^p\right)^{1/p} + \left(\sum_{i=1}^n b_i^p\right)^{1/p}$

This is the generalized triangle inequality $$\|a+b\|_2 \le \|a\|_2 + \|b\|_2$$. It can be deduced immediately using $$g(x)=(x^{1/p}+1)^p$$ with $$f(x,y)=(x^{1/p}+y^{1/p})^p$$, and then substitute $$x_i=a_i^p, y_i=b_i^p$$.

#### Milne inequality

$\left(\sum_{i=1}^n (a_i+b_i)\right)\left(\sum_{i=1}^n \frac{a_ib_i}{a_i+b_i}\right) \le \left(\sum_{i=1}^n a_i\right) \left(\sum_{i=1}^n b_i\right)$

The inequality studied by E. A. Milne and arose from astronomy. This can be obtained by using $$g(x)=x/(1+x)$$ with $$f(x,y)=xy/(x+y)$$.

## Bibliographic data

@article{
title = "When Cauchy and Holder Met Minkowski: A Tour through Well-Known Inequalities",
author = "Gerhard J. Woeginger",
journal = "Mathematics Magazine",
volume = "83",
number = "3",
pages = "202--207",
year = "2009",
}