prove the following inequalities

Let $a_1,a_2,\dots,a_{100}$ be real numbers, each less than one, satisfy

$$a_1 +a_2+ \dots+ a_{100}\gt 1$$

$\mathbf (i)$ Let $n_0$ be the smallest integer $n$ such that $$a_1+a_2+\dots+a_n\gt 1$$

Show that all the sums $\,$ $a_{n_0},a_{n_0}+a_{n_0-1},\dots,a_{n_0}+\dots+a_1$ are positive.

$\mathbf (ii)$ Show that there exist two integers $p$ and $q$, $p\lt q$, such that the numbers

$$a_q,a_q+a_{q-1},\dots,a_q+\dots+a_p$$ $$a_p,a_p+a_{p+1},\dots,a_p+\dots+a_q$$

are all positive.

• I think you made a typo in part $(ii)$, in the second line of displayed math. If $p < q$, then how is it that the sequence $a_{p}, a_{p-1}, \ldots$ can eventually lead to $a_q$?
– 727
Aug 12 '15 at 21:19
• @LJL, Thanks for the observation. It was a typo. It will be $a_p+a_{p+1}$ rather than $a_p+a_{p-1}$. I corrected this.
– user249332
Aug 13 '15 at 5:16

1 Answer

For part $(i)$, suppose to the contrary that $a_{n_0} < 0$. You'll quickly see that the definition of $n_0$ leads you to a contradiction. You can extend this argument to the other sums ...