Two sequences such that $a_i,b_i\in \{-1,0,1\} $ for all $i$ 
Let $(a_i)_{i\in \mathbb{N}}$ and $(b_i)_{i\in \mathbb{N}}$ be two sequences such that :
  $$\forall i\in \mathbb{N}\ \  a_i,b_i\in \{-1,0,1\} $$
Assuming that for all $n\in\mathbb{N^+}$:
  $$a_nb_n+a_n\sum_{i=1}^{n-1}b_i+b_n\sum_{i=1}^{n-1}a_i \in \{-1,0,1\} $$
prove that there exists a constant $c$ such that :
  $$\forall n\in\mathbb{N}: \min\left(\sum_{i=1}^{n}b_i,\sum_{i=1}^{n}a_i\right)\leq c $$

I started proving this by recurrence, but without success. Any suggestions
 A: Let $A_n$ denote $\sum_{i=1}^n a_i$, and $B_n$ similarly.  Notice that
\begin{align}
a_n b_n + a_n \sum_{i=1}^{n-1} b_i + b_n \sum_{i=1}^{n-1} a_i
\end{align}
can be rewritten as $A_n B_n - A_{n-1} B_{n-1}$.  So the given condition just says that $A_i B_i$ never changes by more than one at a time.  Let's rephrase the problem to make it a little more intuitive:  we're given a sequence of lattice points $(A_n, B_n)$ in the plane, starting at the origin and moving at one unit vertically and horizontally at a time, such that the product $A_n B_n$ changes by at most one at each step.  We claim that $A_n$ and $B_n$ are never both less than $-2$, and in fact are never both less than $-1$ unless both of them are equal to $-2$.
By induction, we just have to show that if our claim is true for $n-1$, then it is true for $n$.  To prove this, let's examine how $A_n B_n$ differs from $A_{n-1} B_{n-1}$ in the various cases for the pair $(a_n, b_n)$:
First, we can ignore the cases where neither $a_n$ nor $b_n$ is $-1$, as we have $A_n \geq A_{n-1}$ and $B_n \geq B_{n-1}$, and so $\min(A_n, B_n)$ is no less than $\min(A_{n-1}, B_{n-1})$.
If $a_n = -1$ and $b_n = 0$, then $A_n B_n - A_{n-1} B_{n-1}$ is just $-B_n$.  By the condition we're given, this means that $B_n$ is either $0$ or $\pm 1$, and in particular $B_n > -2$.  So we're fine here.  By symmetry, we're also okay if $a_n = 0$ and $b_n = -1$.
Next, let's consider the case where $a_n = b_n = -1$.  Then we're given that $1 - B_{n-1} -A_{n-1}$ is $0$ or $\pm 1$.  This implies that $A_{n-1} + B_{n-1}$ is nonnegative, so at least one of $A_{n-1}$ and $B_{n-1}$ is nonnegative.  It follows that one of $A_n$ and $B_n$ is at least $-1$, so we're still okay.
Finally (by symmetry again), we're left with the case where $a_n = 1$ and $b_n = -1$.  Then we're given that $-1 + B_{n-1} - A_{n-1}$ is either $0$ or $\pm 1$.  So $A_n = A_{n-1} + 1$ is within $1$ of $B_{n-1}$.  If $A_n \geq B_{n-1}$, then $\min(A_n, B_n) \geq A_n \geq B_{n-1} \geq -2$, so we're fine again.  The only other thing that can happen is that $A_n = B_{n-1} - 1 = B_n$.  This seems like a problem:  it is true that for any $k$, we can move from the point $(-k-1, -k+1)$ to $(-k, -k)$.  But what we have just shown is that moves of this form (and with $a$'s and $b$'s switched) are the only moves possible that end with both $A_n$ and $B_n$ less than $-1$.  If $k > 2$, then we have no way of reaching $(-k-1, -k+1)$ in the first place, as both coordinates are less than $-1$.  So the least possible value of $\min(A_n, B_n)$ that we can achieve is $-2$, which is achieved (for example) via the path
\begin{align}
(0, 0) \to (0, -1) \to (-1, -1) \to (-2, -1) \to (-3, -1) \to (-2, -2).
\end{align}
