I have summarized information about the \(n\)-dimensional Euclidean space \( \mathbb{R}^n \) and its distance below.
(n)-Dimensional Euclidean Space \( \mathbb{R}^n \)
When we define distance as shown below for the direct product set \( \mathbb{R}^n \), we call \( \mathbb{R}^n \) an \(n\)-dimensional Euclidean space. The elements of \( \mathbb{R}^n \) are referred to as points.
Distance
For elements \( x = (x_1, \cdots , x_n) \) and \( y = (y_1, \cdots , y_n) \) in \( \mathbb{R}^n \) (where \( x_i, y_i \in \mathbb{R} \; (i = 1, \cdots , n) \)), their distance \( d(x, y) \) is defined by the following equation:
\[ d(x, y) = \sqrt{\sum _{i=1}^n (x_i – y_i)^2 }\]In 1-dimensional space, \( d(x, y) = | x – y | \), and in 2-dimensional space, \(d(x,y) = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} \).
Properties of Distance
The distance in \( \mathbb{R}^n \) has the following properties:
- For any \(x, y \in \mathbb{R}^n\), \( d(x, y) \) is a non-negative real number.
- For \(x, y \in \mathbb{R}^n\), \( d(x, y) = 0 \) if and only if \( x = y \).
- For any \(x, y \in \mathbb{R}^n\), \( d(x, y) = d(y, x) \).
- For any \( x, y, z \in \mathbb{R}^n \), \( d(x, z) \leq d(x, y) + d(y, z) \) (Triangle Inequality).
The first three properties are evident from the definition of the distance \( d(x, y) \).
Let’s prove the last property, the Triangle Inequality.
Proof
Let \( x = ( x_1, \cdots , x_n) \), \( y = ( y_1 , \cdots , y_n ) \), and \( z = (z_1 , \cdots , z_n ) \). Define \( a_i = x_i – y_i \) and \( b_i = y_i – z_i \). Then, \( d(x, z) \leq d(x, y) + d(y, z) \) can be expressed as follows:
\[ \sqrt{\sum _{i=1}^{n} \left( a_i + b_i \right) ^2} \leq \sqrt{\sum _{i=1}^{n} a_i ^2} + \sqrt{\sum _{i=1}^{n} b_i ^2} \]This equation is equivalent to squaring both sides, eliminating the terms \(\sum a_i^2\) and \(\sum b_i ^2\), and dividing both sides by 2:
\[ \sum _{i=1}^{n} a_i b_i \leq \sqrt{ \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right) } \]Furthermore, this proof is sufficient if we can establish that the squared inequality below is valid:
\[ \left( \sum _{i=1}^{n} a_i b_i \right)^2 \leq \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right) \]This last inequality is known as the Schwarz inequality, and various proofs are known. When \( n = 1 \), equality holds.
Solution 1
Let’s consider the case where \( n \) is greater than or equal to 2. We calculate by subtracting the right side from the left side:
\begin{eqnarray} & & \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right) – \left( \sum _{i=1}^{n} a_i b_i \right)^2 \\ & = & \sum _{\substack{1 \leq i \leq n , 1 \leq j \leq n , i \neq j}} a_i ^2 b_j ^2 – 2 \sum _{1 \leq i \lt j \leq n} a_i b_i a_j b_j \\ & = & \sum _{1 \leq i \lt j \leq n} a_i ^2 b_j ^2 + \sum _{1 \leq i lt j \leq n} a_j ^2 b_i ^2 – 2 \sum _{1 \leq i \lt j \leq n} a_i b_i a_j b_j \\ & = & \sum _{1 \leq i \lt j \leq n} \left( a_i b_j – a_j b_i \right) ^2 \end{eqnarray}Since it is the sum of squares, it is clear that it is non-negative.
The condition for equality to hold is that \( a_i b_j – a_j b_i = 0 \) holds for all distinct \( i, j \).
Consideration of the Condition for Equality
The equation \( a_i b_j – a_j b_i = 0 \) means that when considering 2-dimensional vector spaces \( \mathbb{R}^2 \), \( a = (a_i, a_j) \) and \( b = (b_i, b_j) \), one is a scalar multiple of the other.
This implies that three points \(x, y, z\) lie on the same line. Using vector operations, we can write \( c(x – y) = d(y – z) \), where \( c \) and \( d \) are scalar values.
Let’s solve the equation \( a_i b_j – a_j b_i = 0 \) step by step.
Case \( a_i = 0 \)
This leads to \( a_j = 0 \) or \( b_i = 0 \).
If \( b_i \neq 0 \), then for all \( j \), \( a_j = 0 \), which means \( x = y \
If \( b_i = 0 \), then \( a_i = b_i \) can be written.
Case \( a_i \neq 0 \)
Let \( c = \frac{b_i}{a_i} \).
If \( a_j \neq 0 \), then \( \frac{b_i}{a_i} = \frac{b_j}{a_j} = c \). For all \( i \) where \( a_i \neq 0 \), we have \( c a_i = b_i \ ).
If \( a_j = 0 \), then \( b_j = 0 \), and \( c a_j = b_j \) holds.
It is now clear that the equality holds when \( x, y, z \) are collinear. This also includes the case when \( x = y \).
Solution 2
When \( x = y \), all \( a_i = 0 \), and the inequality holds (equality).
For the case \( x \neq y \), consider \( t \) as a real variable and the quadratic function \( f(t) \):
\begin{eqnarray} f(t) & = & \sum _{i=1}^n \left( a_i t + b_i \right) ^2 \\ & = & \left( \sum _{i=1}^n a_i ^2 \right) t^2 + 2 \left( \sum _{i=1}^n a_i b_i \right) t + \left( \sum _{i=1}^n b_i ^2 \right) \end{eqnarray}From the definition of \( f(t) \) (the first equation), it is clear that \( f(t) \geq 0 \) for any \( t \). This implies that the discriminant is non-negative, leading to the following inequality:
\[ \left( \sum _{i=1}^n a_i b_i \right)^2 – \left( \sum _{i=1}^n a_i ^2 \right) \left( \sum _{i=1}^n b_i ^2 \right) \geq 0 \]By rearranging, it becomes the Schwarz inequality.
Consideration of the Condition for Equality
The condition for equality is when \( f(t) = 0 \) has a double root. Let \( t = t_0 \ ) be that root.
\[ 0 = \sum _{i=1}^n \left( a_i t_0 + b_i \right) ^2 \]At that moment, for all \( i \), \( a_i t_0 + b_i = 0 \). This implies that \( x, y, z \) are vectors and \( – t_0 (x-y) = (y – z) \), meaning that three points \( x, y, z \) are collinear (including the case \( x = y \)).
