No conjectures...

This is an inequality problem I found in a book, but I wonder if it can be done in simple way. Try this:

If a1,a2,,ana_1,a_2, \cdots ,a_n are positive reals less than one and Sn=a1++anS_n = a_1 + \cdots + a_n, then show that

1Sn<(1a1)(1a2)(1an)<11+Sn{1-S_n < (1-a_1)(1-a_2) \cdots (1-a_n) < \displaystyle \frac{1}{1+S_n}}.

Note by A Brilliant Member
5 years, 4 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Also, ai < 1 for all i = 1, 2...n 1 - (a1)^2 < 1 or, (1 - a1)(1+a1) < 1 or, (1- a1) < 1/(1+a1) Multiplying over all terms, (1 - a1)(1- a2)...(1 - an) < 1/(1+a1)(1+a2)...(1+an) But (1+a1)(1+a2)...(1+an) > 1 + Sn This immediately proves it.

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

Again not correct, ai2<1a_i^2 < 1 doesn't imply 1ai2<1 1- a_i^2 < 1. You need to specify proofs more. Anyway for latex, you may use latex editor for Chrome.

A Brilliant Member - 5 years, 4 months ago

Log in to reply

Actually, Since 1>ai2>0 1 > a_i^2 > 0 , we have 1>1ai2>0 1 > 1 - a_i^2 > 0

Siddhartha Srivastava - 5 years, 4 months ago

Log in to reply

@Siddhartha Srivastava Right, but he wrote "or", implying that the statement followed from 1ai2>01- a_i^2 >0. The proof was correct though.

A Brilliant Member - 5 years, 4 months ago

Log in to reply

@A Brilliant Member The OR I wrote was just a continuation :P

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

@A Brilliant Member Like "IMPLIES"

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

@A Brilliant Member And I really don't get you. ai < 1 for all i. So, ai^2 < 1 for all i. This gives, 0 < 1 - ai^2 < 1. Now you just factor out (1- ai^2) into (1 - ai)(1+ai).

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

(1 - a1) (1 - a2) = 1 - a1 - a2 + a1a2 > 1 - (a1 + a2) So, for n such brackets, you can write (1- a1)(1- a2)...(1-an) > 1 - (a1+a2+a3...+an) = 1 - Sn

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

Not rigorous. :( Also there is no chance to justify your claim, in (1a1)(1a2)(1an)(1-a_1)(1-a_2) \cdots (1-a_n), there are many more terms other than (1)na1a2a3an (-1)^n a_1a_2a_3 \cdots a_n. Not correct.

A Brilliant Member - 5 years, 4 months ago

Log in to reply

I think you just missed what I meant to say. I avoided writing entire proofs because I personally don't like answers without LATEX and here I am, without knowing LATEX. Anyway, I have already proved that (1-a1)(1-a2) > 1 - (a1+a2). Then, (1-a1)(1-a2)(1-a3) > (1-a3)(1-a1-a2)>1-(a1+a2+a3). Got it?

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

@Christopher Johnboy Now, this can, similarly be extended to n terms.

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

@Christopher Johnboy I know it would come, my friend, but I thought of getting a rigorous proof. Anyway no problem! :)

A Brilliant Member - 5 years, 4 months ago

Log in to reply

And guys, it would be really really helpful if you can tell me any site where I can write LATEX directly and when I paste it here in comments box, the entire writing with LATEX will come.

Christopher Johnboy - 5 years, 4 months ago

Log in to reply

I use the latex editor Latexian, as it displays the equations as I am typing them.

An online free site would be WriteLatex, which I have used in collaboration with others internationally.

Calvin Lin Staff - 5 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...