2025-02-09

1000: 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

<The previous article in this series | The table of contents of this series | The next article in this series>
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

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:
{r1,...,rk}: ⊆R
//

Statements:
(
∀j∈{1,...,k}(1≤rj)
⟹
r1+...+rk−k+1≤r1...rk
)
∧
(
∀j∈{1,...,k}(2≤rj)
⟹
r1+...+rk−k+1<r1...rk
)
//

2: Note


Especially, rj 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 ∀j∈{1,...,k}(1≤rj) case; Step 2: prove it for the ∀j∈{1,...,k}(2≤rj) case.

Step 1:

Step 1 Strategy: regard the both sides of the inequality as the functions of rk with r1,...,rk−1 regarded as constants, f1(rk):=r1+...+rk−k+1 and f2(rk):=r1...rk, and see that f1(1)≤f2(1) and df1/drk(rk)≤df2/drk(rk).

Let us suppose that ∀j∈{1,...,k}(1≤rj).

Let us think of the k=2 case.

r1+r2−2+1≤r1r2?

Let us regard the both sides as the functions of r2 with r1 regarded as a constant: f1(r2):=r1+r2−2+1 and f2(r2):=r1r2.

f1(1)=r1+1−2+1=r1≤r11=f2(1).

df1/dr2(r2)=1≤r1=df2/dr2(r2), because 1≤r1.

That implies that r1+r2−2+1≤r1r2.

Let us suppose that r1+...+rk−k+1≤r1...rk for k=2,...,l−1.

Let us see that r1+...+rl−l+1≤r1...rl.

Let f1(rl):=r1+...+rl−l+1 and f2(rl):=r1...rl, as before.

f1(1)=r1+...+rl−1+1−l+1=r1+...+rl−1−(l−1)+1≤r1...rl−1=r1...rl−11=f2(1), by the induction hypothesis.

df1/drl(rl)=1≤r1...rl−1=df2/drl(rl).

That implies that r1+...+rl−l+1≤r1...rl.

So, by the induction principle, r1+...+rk−k+1≤r1...rk for each 2≤k.

Step 2:

Step 2 Strategy: regard the both sides of the inequality as the functions of rk with r1,...,rk−1 regarded as constants, f1(rk):=r1+...+rk−k+1 and f2(rk):=r1...rk, and see that f1(2)<f2(2) and df1/drk(rk)<df2/drk(rk).

Let us suppose that ∀j∈{1,...,k}(2≤rj).

Let us think of the k=2 case.

r1+r2−2+1<r1r2?

Let us regard the both sides as the functions of r2 with r1 regarded as a constant: f1(r2):=r1+r2−2+1 and f2(r2):=r1r2.

f1(2)=r1+2−2+1=r1+1<r12=f2(2), because from 1<r1, r1+1<r1+r1=2r1.

df1/dr2(r2)=1<r1=df2/dr2(r2), because 1<r1.

That implies that r1+r2−2+1<r1r2.

Let us suppose that r1+...+rk−k+1<r1...rk for k=2,...,l−1.

Let us see that r1+...+rl−l+1<r1...rl.

Let f1(rl):=r1+...+rl−l+1 and f2(rl):=r1...rl, as before.

f1(2)=r1+...+rl−1+2−l+1=r1+...+rl−1−(l−1)+1+1<r1...rl−1+1<r1...rl−12=f2(2), by the induction hypothesis and from 1<r1...rl−1, r1...rl−1+1<r1...rl−1+r1...rl−1=2r1...rl−1.

df1/drl(rl)=1<r1...rl−1=df2/drl(rl).

That implies that r1+...+rl−l+1<r1...rl.

So, by the induction principle, r1+...+rk−k+1<r1...rk for each 2≤k.

References
<The previous article in this series | The table of contents of this series | The next article in this series>
Posted by at
Labels: