For $ a, b \in \mathbb{Z}^+ $ such that $ a = b $ or $ 2a < b $, let $ f_{a, b} : \mathbb{Z}^+ \to \mathbb{Z}^+ $ be a sequence that satisfies the following three criteria: $$ f(1) = a, \\ f(2) = b, \\ f(n + 2) = f(n + 1) + f(n), n \ge 1. $$ Furthermore, we say that $ f_{a', b'} < f_{a, b} $ iff $ b' < b \vee (b' = b \wedge a' < a ) $.
For a given $ m \in \mathbb{N} $, let $ F_m = \left\{ f_{a, b} : m \in f\left(\mathbb{Z}^+\right) \wedge f^{-1}(m) > 2\right\} $, i.e. the set of sequences that contain a term whose value is $ m $ such that that term is not either of the first two in the sequence. There are exactly $ m - \left\lceil \frac{m}{2}\right\rceil $ elements in $ F_m $, as the term preceding $ m $ lies in $ I_m=\left[\left\lceil \frac{m}{2}\right\rceil, m\right) $.
Because $ f_{a, b} $ is uniquely determined by any two consecutive terms, we can define a new function $ g_m $ such that $ g_m(n) $ is the unique sequence $ f_{a_{m, n}, b_{m, n}} $ containing consecutive terms $ n $ and $ m $. This is well defined only for $ n \in I_m $.
Prove that there exists integer $ x_m \in I_m $ such that $ g_m $ is decreasing on $ \left[\left\lceil \frac{m}{2}\right\rceil, x\right] $ and increasing on $ [x, m) $.
For example, consider the case $ m = 8 $. The four sequences in $ F_8 $ are $$g_8(4) = f_{4, 4} > g_8(5) = f_{1, 1} < g_8(6) = f_{2, 2} < g_8(7) = f_{1, 7}, $$ and $ x_8 = 5 $.
Also, if possible, find a closed form expression for $ x_m $ in terms of $ m $, although it seems unlikely that this exists.