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 $a_1,a_2, \cdots ,a_n$ are positive reals less than one and $S_n = a_1 + \cdots + a_n$, then show that

${1-S_n < (1-a_1)(1-a_2) \cdots (1-a_n) < \displaystyle \frac{1}{1+S_n}}$ .

#Goldbach'sConjurersGroup #Inequality

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Comments

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 - 7 years, 1 month ago

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

A Brilliant Member - 7 years, 1 month ago

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

Siddhartha Srivastava - 7 years, 1 month ago

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

A Brilliant Member - 7 years, 1 month ago

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

Christopher Johnboy - 7 years, 1 month ago

@A Brilliant Member – Like "IMPLIES"

Christopher Johnboy - 7 years, 1 month ago

@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 - 7 years, 1 month ago

(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 - 7 years, 1 month ago

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

A Brilliant Member - 7 years, 1 month ago

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 - 7 years, 1 month ago

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

Christopher Johnboy - 7 years, 1 month ago

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

A Brilliant Member - 7 years, 1 month ago

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 - 7 years, 1 month ago

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 - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$