Solve inequality $F(z) \leq\min(1, z) $

Question

I am struggling to solve this equality that leads to the acceptance probability of a Markov chain:

$$ \frac{F(z)}{F(1/z)} = z $$

where $F(z) < 1$ and $F(1/z) < 1$. So apparently I get $$ F(z) < z $$ and later $$ F(z) \leq \min(1, z).$$ Any ideas?

Answer 1

ok I solved it:

$$ F(z) = F(1/z) \cdot z \leq z $$

because $ F(1/z) < 1$.