There are two functions $f:\mathbb N\to \mathbb R^+$ and $g:\mathbb N\to \mathbb R^+$. For all $n\in \mathbb N$: $$f(1)=1$$ $$g(1)=2$$ $$f(n+1)=\frac{1+f(n)+f(n)g(n)}{g(n)}$$$$g(n+1)=\frac{1+g(n)+f(n)g(n)}{f(n)}$$ Prove that $f(2020)<5$.
Notes:
- It appears that $\lim_{x\to \infty}f(x)=5$ and $\lim_{x\to \infty}g(x)=\infty$, though I haven't been able to prove either of these statements.
- First few results: $f(1)=1; f(2)=2; f(3)=2.6; f(4)=3.05$ and $g(1)=2; g(2)=5; g(3)=8; g(4)=11.\overline{461538}$.
We have $f(n+1) + 1= \frac{(f(n)+1)(g(n)+1)}{g(n)}$ and for $g(n+1)$ a similar equation. With this we get $$\frac{1}{f(n+1)+1}-\frac{1}{g(n+1)+1} = \frac{1}{f(n)+1}-\frac{1}{g(n)+1} = \frac{1}{f(1)+1}-\frac{1}{g(1)+1} =\frac{1}{6}$$ and therefore $ \frac{1}{f(n)+1} = \frac{1}{6} + \frac{1}{g(n)+1}>\frac{1}{6}\Rightarrow f(n)<5$.
The solution is short but it doesn't explain much. :/