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

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

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## Comments

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TopNewestAlso, 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.

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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.

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Actually, Since \( 1 > a_i^2 > 0 \), we have \( 1 > 1 - a_i^2 > 0 \)

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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.

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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.

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(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

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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.

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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?

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rigorousproof. Anyway no problem! :)Log in to reply