let two positive sequence $\begin{cases} x_{n+2}=x_{n}+x^2_{n+1}\\ y_{n+2}=y^2_{n}+y_{n+1} \end{cases}$ and $x_{1}>1,y_{1}>1,x_{2}>1,y_{2}>1$
show that: there exists $n$, such $x_{n}>y_{n}$
my idea: since
$$x_{n+2}-y_{n+2}=(x_{n}-y_{n+1})+(x^2_{n+1}-y^2_{n})$$ then follow I can't have idea.
maybe this problem can use contradiction
we Assume that $$x_{n}\le y_{n}\forall n\in N^{+}$$
A year ago, I have post this How prove this $|x_{p}-y_{q}|>0$
But this two different problem,maybe can use same methods?
Prove by induction that $y_{n+2} < Cy_{n+1}^{3/2}$ for some positive constant $C$. Indeed, the $C$ is only needed in the base case. (Motivation: each term in the $y$ sequence is approximately the square of the term two terms earlier, once they become large; so maybe each term is like the $\sqrt2$ power of the previous term, and $\sqrt2<\frac32$.) Together with $x_{n+2}>x_{n+1}^2$, it's easy to deduce the desired conclusion.