Finding the maximum amount of positive numbers in a sequence Let $(a_1,...,a_{100})$ be real numbers satisfying :
$\forall i\in\mathbb{Z}/100\mathbb{Z}$, $a_i > a_{i+1}+a_{i+2}$. What is the maximum amount of strictly positive real numbers contained in the sequence $(a_i)_{1\le i\le100}$.
The obvious issue is that if you wish for all the reals to be positive, then they must be decreasing and so when you loop around you'll meet a problem. I feel like it should be possible to alternate between positive and negative numbers, but I'm not sure how to show this.
 A: This is essentially taken from Combinatorics on AoPS.
There can be at most $50$ positive numbers because otherwise a there would be two positive numbers $a_{i+1}, a_{i+2}$ immediately following a non-positive number $a_i$.
If there are $50$ positive numbers then the $a_i$ must be alternatingly positive and non-positive. Without loss of generality assume that $a_i > 0$ for odd $i$. Then $a_i > a_{i+2}$ for even $i$ and
$$
 a_{100} > a_2 > a_4 > \ldots > a_{98} > a_{100}
$$
gives a contradiction.
So there can be at most $49$ strictly positive numbers. An example with $49$ positive numbers is
$$
(-101, 1, -103, 1, -105, 1, \ldots, -197, 1, -199, -99)
$$
i.e.
$$
 a_i = \begin{cases}
 -100-i & \text{ if $i$ is odd,} \\
 +1 & \text{ if $i < 100 $ is even,} \\
 -99 & \text{ if $i = 100$.} 
\end{cases}
$$

In the same way one can show that if $n$ is even and $(a_1, a_2, \ldots, a_n)$ are real numbers satisfying $a_i > a_{i+1}+a_{i+2}$ for all $i$ (with wrapping around indices) then there are at most $n/2-1$ strictly positive numbers among these.
An example with $n/2-1$ strictly positive numbers is
$$
 (-n-1, 1, -n-3, 1, -n-5, 1, \ldots, -2n+3, 1, -2n+1, -n+1)
$$
i.e.
i.e.
$$
 a_i = \begin{cases}
 -n-i & \text{ if $i$ is odd,} \\
 +1 & \text{ if $i < n $ is even,} \\
 -n+1 & \text{ if $i = n$.} 
\end{cases}
$$
