Clipping a negative prefix followed by a positive suffix

Put $s_j=x_0+{\sum }_{i=1}^j{\delta }_i$ and $s_0=x_0$. For $0\le j\le k$, induction gives $x_j\ge s_j$ and $x_j\le 1$: the base is $x_0=s_0\in [-1,1]$; at a next nonpositive increment, $x_{j-1}+{\delta }_j\le 1$, so clipping replaces it only by its maximum with $-1$, which is no smaller than $x_{j-1}+{\delta }_j\ge s_j$.

For $k\le j\le n$, induction gives $x_j\ge \min \{1,s_j\}$. The base follows from step 1.1 and $x_k\le 1$. If $s_{j-1}\le 1$, then $x_{j-1}\ge s_{j-1}$ and ${\delta }_j\ge 0$, so $y:=x_{j-1}+{\delta }_j\ge s_j$. If $s_j<-1$, then $c(y)\ge -1>s_j$. If $-1\le s_j\le 1$, both $1$ and $y$ are at least $s_j$, so $\min \{1,y\}\ge s_j$ and then $c(y)=\max \{-1,\min \{1,y\}\}\ge s_j$. If $s_j>1$, then $y>1$ and $c(y)=1$. In all cases $c(y)\ge \min \{1,s_j\}$. If instead $s_{j-1}>1$, the induction bound and $x_{j-1}\le 1$ imply $x_{j-1}=1$; a nonnegative increment keeps the clipped value at $1$. Taking $j=n$ proves the Statement.