Solving a recurrence relation with variable coefficients

I am trying to solve this recurrence inequality for $$a_n$$, hopefully in terms of $$a_0，a_1，c$$ and $$x$$,

$$(1-x)a_{n+1} \le xa_n + (1+x)a_{n-1} + c\quad\forall\ n\ge1$$

What I tried: RSolve. But RSolve expects an equality, and not an inequality. Any tips on how to proceed? $$a_i \in \mathbb{R}$$, $$x\in\mathbb{Q}$$, $$c > 0$$, $$0 < x < 1$$ and $$a_i$$ is not constant for all $$i$$.

Edit: I have some control over $$x$$ so I chose $$x=1/2$$ and used an equality instead of an inequality, then applied RSolve.

Simplify[RSolve[(1/2) a[n] == (1/2) a[n - 1] + (3/2)*a[-2 + n] + c, a[n], n]]

I got the result $$a(n)\to c_1 \left(\frac{1}{2} \left(1-\sqrt{13}\right)\right)^n+c_2 \left(\frac{1}{2} \left(\sqrt{13}+1\right)\right)^n-\frac{2 c}{3}.$$ However what I need is the inequality, and in terms of $$x$$. Do you know any other method in Mathematica available to do this?

• As the error message says, RSolve is expecting an equation. You gave it an inequality. Change <= to == and it will return a result. – Carl Woll Oct 29 '18 at 2:37
• Hi, I am open to other ways to solve the recurrence without RSolve. I have searched but I can't find the functions I need. – Cogicero Oct 29 '18 at 2:38
• Thanks Bill, that works, but my problem is that $a$ is definitely not constant for all $n$ in my case. – Cogicero Oct 29 '18 at 3:14
• Thanks. This is one step in some numerical analysis. I know that $a$ depends on $n$ i.e. $a_i$ is not constant for all $i$. I know that $x > 0$ and $c > 0$ while $n >= 1$. $x \in \mathbb{Q}$ while $a_i \in \mathbb{R}$ and $c \in \mathbb{R}$. – Cogicero Oct 29 '18 at 3:25
• I don't understand the problem yet. Let's take a simple special case, say a0=0 and a1=1. What kind of answer would you expect to see? – bill s Oct 29 '18 at 14:41