Let $\mathbb N$ be the set of naturals, and let $g:\mathbb N\times \mathbb N \to [0,\infty)$ be a function. Define, for all natural numbers, $$f(n)=\frac{a g(n-1,n)+b g(n,n+1)}{c g(n-1,n)+d g(n,n+1)+ e}$$ For what positive real numbers $a, b, c, d, e$ is the sequence $\{f(n)\}$ non-increasing?
So I need to prove that for all $n=1,2,3...$, $f(n)\geq f(n+1)$.
For a special case $a=b=c=d=e=1$, we see $\{f(n)\}$ nonincreasing.
I am trying to evaluate $f(n)-f(n+1)\geq 0$ on paper, but unable to to do the next step.
Can we consider this function $$f(x)=\frac{a g(x-1,x)+b g(x,x+1)}{c g(x-1,x)+d g(x,x+1)+ e}$$ and determine the values of parameters for which $f'(x)\leq 0$? Derivative is given here