# 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

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

Note by Paramjit Singh
3 years, 12 months ago

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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.

- 3 years, 12 months 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.

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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).

- 3 years, 12 months ago

Like "IMPLIES"

- 3 years, 12 months ago

The OR I wrote was just a continuation :P

- 3 years, 12 months 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.

- 3 years, 12 months 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.

Staff - 3 years, 12 months 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

- 3 years, 12 months 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.

- 3 years, 12 months 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?

- 3 years, 12 months ago

Now, this can, similarly be extended to n terms.

- 3 years, 12 months ago

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

- 3 years, 12 months ago