Example 4 Suppose
$$
x=\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}+\frac{1+1}{1+\cdots}}}
$$

with 1000 horizontal fraction bars. Is $x^{2}+x>1$ always true? (1990, Wisconsin Talent Search)
Let $x_{n}$ represent a number in the form of $x$ with $n$ fraction lines. Then, we have $x_{n+1}=\frac{1}{1+x_{n}}$.
It is easy to find $x_{1}=\frac{1}{2}, x_{2}=\frac{2}{3}, x_{3}=\frac{3}{5}, x_{4}=\frac{5}{8}$,
$\cdots$. Let $y_{n}=x_{n}^{2}+x_{n}$, then
$$
y_{1}=\frac{3}{4}1, y_{3}=\frac{24}{25}1$. Conjecture: when $n$ is odd, $y_{n}1$. We will prove this conjecture.
$$
\begin{aligned}
\because & y_{n+1}-1=x_{n+1}^{2}+x_{n+1}-1 \\
& =\left(\frac{1}{1+x_{n}}\right)^{2}+\left(\frac{1}{1+x_{n}}\right)-1 \\
& =\left(\frac{1}{1+x_{n}}\right)^{2}\left[1+\left(1+x_{n}\right)-\left(1+x_{n}\right)^{2}\right] \\
& =-\left(\frac{1}{1+x_{n}}\right)^{2}\left(x_{n}^{2}+x_{n}-1\right) \\
& =-\left(\frac{1}{1+x_{n}}\right)^{2}\left(y_{n}-1\right),
\end{aligned}
$$
$\therefore y_{n+1}-1$ has the opposite sign of $y_{n}-1$, which means the conjecture is true.
Thus, $y_{1000}>1$.
$$
\therefore x^{2}+x>1 \text {. }
$$