Show that $x_i\leq \max{x_0,x_n,0}$, discrete maximum principle

Question

Let $x=(x_0,...,x_n)\in \mathbb R^{n+1}$that satisfies $a_ix_{i-1}+b_ix_i+c_ix_{i+1}<0$ with coefficients $a_i,b_i,c_i\in\mathbb R$ with $a_i,c_i<0, b_i>0, a_i+b_i+c_i\geq 0$ for $1\leq i< n$. Show that $x_i\leq \max\{x_0,x_n,0\}$ for all $i\in\{0,...,n\}$.
I tried to prove it by contradiction. This means there exists $i\in\{1,...,n-1\}$ so that $x_i\geq \max\{x_{i-1},x_{i+1},0\}$ and $x_i\geq \min\{x_{i-1},x_{i+1}\}$. Can someone give me a hint?

Answer 1

Let $$ \tau:= \max\{x_i: ~ i \in \{0,\dots,n\}\}. $$ Then there exists (at leat one) $k\in \{0,\dots,n\}$ such that $\tau= x_k$.

1. case: $k\in \{0,n\}$. Then $$ \forall i \in \{0,\dots,n\}: ~ x_i \le \tau = x_k = \max\{x_0,x_n\} \le \max\{x_0,x_n,0\}. $$
2. case: $k\in \{1, \dots,n-1\}$. Then $\tau \ge x_{k-1}$ and $\tau \ge x_{k+1}$, hence $$ 0 > a_kx_{k-1}+b_kx_k+c_kx_{k+1}= a_kx_{k-1}+b_k\tau+c_kx_{k+1} \ge a_k\tau+b_k\tau+c_k\tau = (a_k+b_k+c_k)\tau. $$ Since $a_k+b_k+c_k \ge 0$ we get $\tau < 0$, hence $$ \forall i \in \{0,\dots,n\}: ~ x_i \le \tau < 0 \le \max\{x_0,x_n,0\}. $$