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Let $a_1,a_2,\ldots,a_{100}$ be real numbers,each less than one,satisfy $$a_1+a_2+\ldots+a_{100}>1$$ Prove the following statements:
(i) Let $n_0$ be the smallest integer $n$ such that $$a_1+a_2+\ldots+a_n>1$$
Show that all the sums $a_{n_0},a_{n_0}+a_{n_0-1},\ldots,a_{n_0}+\ldots+a_1$ are positive.
(ii) Show that there exist two integers $p$ and $q,p<q$, such that the numbers $$a_q,a_q+a_{q-1},\ldots,a_q+\ldots+a_p$$ $$a_p,a_p+a_{p+1},\ldots,a_p+\ldots+a_q$$
are all positive.

My work:
I could solve the first part by the Extremal Principle, but cannot approach the second part. I do not know extremal principal much. Please help!

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  • $\begingroup$ @Arash...where did you edit? $\endgroup$ – Hawk Jan 27 '14 at 16:31
  • $\begingroup$ Just at the beginning where you put $a_100$. :) $\endgroup$ – Arash Jan 27 '14 at 16:36
  • $\begingroup$ but...didn't I rectify it? Anyways...must have missed it...thanks! $\endgroup$ – Hawk Jan 27 '14 at 16:37
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(ii):

Let $s_0$ be the largest integer $s$ (less than $n_0$) s.t.

$a_s + a_{s+1} + \dots + a_{n_0} > 1$

Then you can take $p=s_0$ and $q=n_0$ since if any of the sums in question is nonpositive it can be discarded to tighten the bounds (and the bounds were chosen "extremaly"). Also $p<q$ since the numbers $a_i$ are less than one.

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