description/proof of that for \(2\) \(n\)-sequences of real numbers, square-root of sum of squares of sums of corresponding items of sequences is equal to or smaller than sum of square-roots of sums of squares of items of sequences
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of sequence.
- The reader knows a definition of Euclidean inner product on Euclidean vectors space.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) \(n\)-sequences of real numbers, the square-root of the sum of the squares of the sums of the corresponding items of the sequences is equal to or smaller than the sum of the square-roots of the sums of the squares of the items of the sequences.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(s_1\): \(: \{1, ..., n\} \to \mathbb{R}\)
\(s_2\): \(: \{1, ..., n\} \to \mathbb{R}\)
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Statements:
\(\sqrt{\sum_{j \in \{1, ..., n\}} (s_1 (j) + s_2 (j))^2} \le \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} + \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\)
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2: Proof
Whole Strategy: Step 1: see that \(\sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \le \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\); Step 2: see that \(\sum_{j \in \{1, ..., n\}} (s_1 (j) + s_2 (j))^2 \le (\sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} + \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2})^2\); Step 3: conclude the proposition.
Step 1:
Let us think of the Euclidean vectors space, \(\mathbb{R}^n\), with the Euclidean inner product.
Let us think of \((s_1 (1), ..., s_1 (n)), (s_2 (1), ..., s_2 (n)) \in \mathbb{R}^n\).
\(\vert \langle (s_1 (1), ..., s_1 (n)), (s_2 (1), ..., s_2 (n)) \rangle \vert \le \sqrt{\langle (s_1 (1), ..., s_1 (n)), (s_1 (1), ..., s_1 (n)) \rangle} \sqrt{\langle (s_2 (1), ..., s_2 (n)), (s_2 (1), ..., s_2 (n)) \rangle}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
That means that \(\vert \sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \vert \le \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j) s_1 (j)} \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j) s_2 (j)}\).
As \(\sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \le \vert \sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \vert\), \(\sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \le \vert \sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \vert \le \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\).
Step 2:
\(2 \sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \le 2 \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\).
\(\sum_{j \in \{1, ..., n\}} s_1 (j)^2 + \sum_{j \in \{1, ..., n\}} s_2 (j)^2 + 2 \sum_{j \in \{1, ..., n\}} s_1 (j) s_2 (j) \le \sum_{j \in \{1, ..., n\}} s_1 (j)^2 + \sum_{j \in \{1, ..., n\}} s_2 (j)^2 + 2 \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\).
The left hand side is \(\sum_{j \in \{1, ..., n\}} (s_1 (j) + s_2 (j))^2\).
The right hand side is \((\sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} + \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2})^2\).
So, \(\sum_{j \in \{1, ..., n\}} (s_1 (j) + s_2 (j))^2 \le (\sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} + \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2})^2\).
Step 3:
So, \(\sqrt{\sum_{j \in \{1, ..., n\}} (s_1 (j) + s_2 (j))^2} \le \sqrt{\sum_{j \in \{1, ..., n\}} s_1 (j)^2} + \sqrt{\sum_{j \in \{1, ..., n\}} s_2 (j)^2}\).
