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Let $ a_1,a_2,\ldots ,a_{100}$ be real numbers, each less than one, satisfy $ a_1+a_2+\cdots+a_{100} > 1$

Show that there exist two integers $p$ and $q$ , $p<q$, such that the numbers

$$a_q, a_q+a_{q-1}, \ldots, a_q+\cdots+a_p,$$

$$a_p, a_p+a_{p+1},\ldots,a_p+\cdots+a_q$$

are all positive.

I proved that if $n$ is the smallest integer such that $ a_1+a_2+\cdots+a_n>1$ then all the sums $a_n,a_n+a_{n-1},\ldots,a_n+\cdots+a_1$ are positive. Will this help to prove?

source: Test of Math at 10+2 level( A collection of old ISI B.stat & B.math entrance exam question papers)

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  • $\begingroup$ Can someone suggest me a better tag for the question? $\endgroup$ – Akshay Hegde Nov 30 '15 at 4:53
  • $\begingroup$ Your tags are fine. The question is really a combinatorics question (and not quite an inequality question), and it indeed sounds like contest-math. However, it is best to clearly state where you got the problem from, so that we know it's not an ongoing contest (otherwise the question is not allowed). $\endgroup$ – user21820 Nov 30 '15 at 5:16
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[I found this solution collaboratively with someone else offline.] $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Let $T(n) = ( \text{The theorem is true for any length-$n$ sequence from $\rr$} )$, for any $n \in \nn$.

If $T(n)$ is false for some $n \in \nn$:

  Let $m \in \nn$ be the minimum such that $T(m)$ is false [by well-ordering].

  Let $a_{1..m}$ be a sequence that does not satisfy the theorem.

  For any $p \in [1..m]$:

    If $\sum_{k=1}^p a_k \le 0$:

      $\sum_{k=p+1}^m a_k \ge \sum_{k=1}^m a_k > 1$.

      Also $a_{p+1..m}$ satisfies the theorem [by minimality of $m$].

      Thus some segment of $a_{p+1..m}$ has initial and terminal segments all with positive sum.

      But any segment of $a_{p+1..m}$ is also a segment of $a_{1..m}$.

      Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.

    Therefore $\sum_{k=1}^p a_k > 0$.

    Similarly $\sum_{k=p}^m a_k > 0$.

  Therefore $a_{1..m}$ has initial and terminal segments all with positive sum.

  Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.

Therefore $T(n)$ is true for any $n \in \nn$.

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  • $\begingroup$ I didn't get some parts of the answer: 1) $a_{1..m} $for a sequence? 2) $\sum ^{p}_{k=1} \leq 0$? $\endgroup$ – Akshay Hegde Nov 30 '15 at 5:03
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    $\begingroup$ @AkshayHegde: (1) $a_{1..m}$ is just a short-hand for $a_1,a_2,\cdots,a_m$. Same for $[1..m]$ being a short-hand for the collection of all integers from $1$ to $m$. (2) Sorry I missed out the expression to be summed. Edited. $\endgroup$ – user21820 Nov 30 '15 at 5:13
  • $\begingroup$ @AkshayHegde: So do you get it? $\endgroup$ – user21820 Dec 1 '15 at 7:44
  • $\begingroup$ Yeah got it... struggled initially though @user21820 $\endgroup$ – Akshay Hegde Dec 1 '15 at 9:24
  • $\begingroup$ @AkshayHegde: Ah okay great. In general, considering the smallest counter-example (where size is a natural number) can help, even though well-ordering is equivalent to induction. This is because when attempting a proof by contradiction you would start with existence of a counter-example, which by well-ordering immediately gives existence of a smallest counter-example, which is often a lot more information to work with. $\endgroup$ – user21820 Dec 1 '15 at 10:16
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If the sum is positive, at least one element is positive. Pick that one as "sequence" (of length 1).

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  • $\begingroup$ Please elaborate.. I didn't understand $\endgroup$ – Akshay Hegde Dec 26 '15 at 3:32

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