description/proof of that for finite number of real numbers equal to or larger than 1 (2), sum of real numbers minus finite number plus 1 is equal to or smaller (just smaller) than product of real numbers
Topics
About: analysis
The table of contents of this article
Starting Context
- The reader knows a definition of partial derivative of function.
Target Context
- The reader will have a description and a proof of the proposition that for any finite number of any real numbers equal to or larger than 1 (2), the sum of the real numbers minus the finite number plus 1 is equal to or smaller (just smaller) than the product of the real numbers.
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:
\(\{r_1, ..., r_k\}\): \(\subseteq \mathbb{R}\)
//
Statements:
(
\(\forall j \in \{1, ..., k\} (1 \le r_j)\)
\(\implies\)
\(r_1 + ... + r_k - k + 1 \le r_1 ... r_k\)
)
\(\land\)
(
\(\forall j \in \{1, ..., k\} (2 \le r_j)\)
\(\implies\)
\(r_1 + ... + r_k - k + 1 \lt r_1 ... r_k\)
)
//
2: Note
Especially, \(r_j\) s can be taken to be some natural numbers.
3: Proof
Whole Strategy: prove it inductively with respect to \(k\); Step 1: prove it for the \(\forall j \in \{1, ..., k\} (1 \le r_j)\) case; Step 2: prove it for the \(\forall j \in \{1, ..., k\} (2 \le r_j)\) case.
Step 1:
Step 1 Strategy: regard the both sides of the inequality as the functions of \(r_k\) with \(r_1, ..., r_{k - 1}\) regarded as constants, \(f_1 (r_k) := r_1 + ... + r_k - k + 1\) and \(f_2 (r_k) := r_1 ... r_k\), and see that \(f_1 (1) \le f_2 (1)\) and \(d f_1 / d r_k (r_k) \le d f_2 / d r_k (r_k)\).
Let us suppose that \(\forall j \in \{1, ..., k\} (1 \le r_j)\).
Let us think of the \(k = 2\) case.
\(r_1 + r_2 - 2 + 1 \le r_1 r_2\)?
Let us regard the both sides as the functions of \(r_2\) with \(r_1\) regarded as a constant: \(f_1 (r_2) := r_1 + r_2 - 2 + 1\) and \(f_2 (r_2) := r_1 r_2\).
\(f_1 (1) = r_1 + 1 - 2 + 1 = r_1 \le r_1 1 = f_2 (1)\).
\(d f_1 / d r_2 (r_2) = 1 \le r_1 = d f_2 / d r_2 (r_2)\), because \(1 \le r_1\).
That implies that \(r_1 + r_2 - 2 + 1 \le r_1 r_2\).
Let us suppose that \(r_1 + ... + r_k - k + 1 \le r_1 ... r_k\) for \(k = 2, ..., l - 1\).
Let us see that \(r_1 + ... + r_l - l + 1 \le r_1 ... r_l\).
Let \(f_1 (r_l) := r_1 + ... + r_l - l + 1\) and \(f_2 (r_l) := r_1 ... r_l\), as before.
\(f_1 (1) = r_1 + ... + r_{l - 1} + 1 - l + 1 = r_1 + ... + r_{l - 1} - (l - 1) + 1 \le r_1 ... r_{l - 1} = r_1 ... r_{l - 1} 1 = f_2 (1)\), by the induction hypothesis.
\(d f_1 / d r_l (r_l) = 1 \le r_1 ... r_{l - 1} = d f_2 / d r_l (r_l)\).
That implies that \(r_1 + ... + r_l - l + 1 \le r_1 ... r_l\).
So, by the induction principle, \(r_1 + ... + r_k - k + 1 \le r_1 ... r_k\) for each \(2 \le k\).
Step 2:
Step 2 Strategy: regard the both sides of the inequality as the functions of \(r_k\) with \(r_1, ..., r_{k - 1}\) regarded as constants, \(f_1 (r_k) := r_1 + ... + r_k - k + 1\) and \(f_2 (r_k) := r_1 ... r_k\), and see that \(f_1 (2) \lt f_2 (2)\) and \(d f_1 / d r_k (r_k) \lt d f_2 / d r_k (r_k)\).
Let us suppose that \(\forall j \in \{1, ..., k\} (2 \le r_j)\).
Let us think of the \(k = 2\) case.
\(r_1 + r_2 - 2 + 1 \lt r_1 r_2\)?
Let us regard the both sides as the functions of \(r_2\) with \(r_1\) regarded as a constant: \(f_1 (r_2) := r_1 + r_2 - 2 + 1\) and \(f_2 (r_2) := r_1 r_2\).
\(f_1 (2) = r_1 + 2 - 2 + 1 = r_1 + 1 \lt r_1 2 = f_2 (2)\), because from \(1 \lt r_1\), \(r_1 + 1 \lt r_1 + r_1 = 2 r_1\).
\(d f_1 / d r_2 (r_2) = 1 \lt r_1 = d f_2 / d r_2 (r_2)\), because \(1 \lt r_1\).
That implies that \(r_1 + r_2 - 2 + 1 \lt r_1 r_2\).
Let us suppose that \(r_1 + ... + r_k - k + 1 \lt r_1 ... r_k\) for \(k = 2, ..., l - 1\).
Let us see that \(r_1 + ... + r_l - l + 1 \lt r_1 ... r_l\).
Let \(f_1 (r_l) := r_1 + ... + r_l - l + 1\) and \(f_2 (r_l) := r_1 ... r_l\), as before.
\(f_1 (2) = r_1 + ... + r_{l - 1} + 2 - l + 1 = r_1 + ... + r_{l - 1} - (l - 1) + 1 + 1 \lt r_1 ... r_{l - 1} + 1 \lt r_1 ... r_{l - 1} 2 = f_2 (2)\), by the induction hypothesis and from \(1 \lt r_1 ... r_{l - 1}\), \(r_1 ... r_{l - 1} + 1 \lt r_1 ... r_{l - 1} + r_1 ... r_{l - 1} = 2 r_1 ... r_{l - 1}\).
\(d f_1 / d r_l (r_l) = 1 \lt r_1 ... r_{l - 1} = d f_2 / d r_l (r_l)\).
That implies that \(r_1 + ... + r_l - l + 1 \lt r_1 ... r_l\).
So, by the induction principle, \(r_1 + ... + r_k - k + 1 \lt r_1 ... r_k\) for each \(2 \le k\).
