$ f_{n + 2}\leq \max\{f_{n},f_{n + 1}\}$ So $ \{f_{n}\}$ is bound . Exist $ k > l$ such that $ f_{l + 1} = f_{k + 1},f_l = f_k$ So $ f_{n + k - l} = f(n)$ for n large enough.

Are you sure it's a $ \ge$ there? Also, you have to prove it becomes constant, while I think you only proved it is periodical.

$ f_{n + 2}\leq \max\{f_{n},f_{n + 1}\}$ So $ f(n)$ is bound. If $ f(n),f(n + 1)$ is diferent then $ f(n + 2)\leq max\{f(n),f(n + 1)\} - 1$ $ f(n) > 0,\forall n$ So when n large enough we must has $ f(n) = f(n + 2) = f(n + 1)$ It mean that for n large enough then it constant.

Let $ gcd(a,b)=d*2^k$, were d is odd, then $ d|f_n$ and for $ n>2log_2(max(\frac{a}{d},\frac{b}{d}))$ we get $ f_n=d.$

Step 1:Show $ f_n$ is constant for sufficiently large $ n$(see post $ 4$) Step 2:It is easy enough to show $ (f_n,f_{n+1})=(f_{n+1},f_{n+2})$ Then the constant is $ (a,b)$