\(\mathbb{R}\) equipped with total order:
\[ \left\{\begin{array}{ll} a < b, ~~~~ &\text{if} ~ a - b ~ \text{is negative} \\ a = b, ~~~~ &\text{if} ~ a - b = 0 \\ a > b, ~~~~ &\text{if} ~ a - b ~ \text{is positive} \end{array}\right. \]
That is, the comparison between real numbers should be done by subtraction.
Example. Which one is larger: \((x+3)(x+7)\) or \((x+4)(x+6)\)
Do Subtraction:
\[ (x+3)(x+7) - (x+4)(x+6) = (x^2+10x+21)-(x^2+10x+24) = -3 < 0 \]
which shows that \((x+3)(x+7) < (x+4)(x+6)\).
As a shortcut to see this result at a glance, we notice the given polynomial fits the form as formula of square difference
\[ (x+\Delta x)(x-\Delta x) = x^2 - (\Delta x)^2 > x^2 \]
This implies that as the offset \(\Delta x\) increased, the result will be smaller.
The properties of inequality are listed following without prove, since it’s easy to give the proof based on the definition:
\(a < b \Leftrightarrow b > a\).
Transitivity. \(a < b, b < c \Rightarrow a < c\).
First-order operation. \(a < b \Rightarrow a+c < b+c\).
Furthermore, use this property and the transitivity, we can prove that, \(a > b, c > d \Rightarrow a + c > b + d\), which described as “The inequality with the same direction can be added together.”
In another perspective, it can be also regarded as \(a > b, d < c \Rightarrow a - d > b - d\), described as “The inequality with the different direction can be subtracted.”
These two asserts is useful thought-shortcut during comparison.
Second-order operation. If \(a < b\), and \(c\) is a positive number, then \(ac > bc\). Otherwise, if \(c\) is a negative number, then \(ac < bc\).
This is a property that strongly different with the equation. The negative number multiplication changes the direction of inequality.
Multiple the inequality \(a < b\) by itself \(n-1\) times, we got \(a^n < b^n ~~ (n\in\mathbb{N}^+)\). With the contradiction proof technique applied, we can also prove that \(\sqrt[n]{a} < \sqrt[n]{b} ~~ (n\in\mathbb{N}^+)\).
The experience tells us that, if you added sugar into the syrup, it will be sweeter, which indicates that
\[ a > b \Rightarrow \cfrac{a+m}{b+m} > \cfrac{a}{b}, ~~~~ m > 0 \]
This can be easily proved by subtraction:
\[ \cfrac{a+m}{b+m} - \cfrac{a}{b} = \cfrac{b(a+m)-a(b+m)}{b(b+m)} = \cfrac{m}{b(b+m)}(b-a) > 0 \]
which can be described as “Add the same positive number to the denominator and numerator in the proper fraction, the value of fraction will be larger.”
Besides, if \(a > b\), which means the fraction is an improper fraction, then
\[ a < b \Rightarrow \cfrac{a+m}{b+m} < \cfrac{a}{b}, ~~~~ m > 0 \]
The application of syrup inequality is gathered in Application of Syrup Inequality.
The square sum inequality based on the non-negative property of square:
\[ (a-b)^2 \ge 0 \Rightarrow a^2 + b^2 \ge 2ab \]
Both sides are equal if and only if \(a = b\). This inequality can be regarded as a various (or the proof middle result) of the famous AM-GM inequality.