Example 7 Given $a \neq b$, and
$$
a^{2}-4 a+1=0, b^{2}-4 b+1=0 \text {. }
$$

Find the value of $\frac{1}{a+1}+\frac{1}{b+1}$.
Solution 1: Since
$$
\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1},
$$

we only need to find the values of $a b$ and $a+b$.
From the given conditions, $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so
$$
a+b=4, a b=1 \text {. }
$$

Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1}$
$$
=\frac{4+2}{1+4+1}=1 \text {. }
$$

This solution has been published in many newspapers.
Analysis: Although the analysis for solving the problem is quite reasonable, and it is necessary to find $a+b$ and $a b$, the conclusion is entirely correct. However, from the review of the solution, to get
$$
\frac{(a+b)+2}{a b+(a+b)+1}=1,
$$

having $a b=1$ is sufficient, and finding $a+b$ is an extra step. Therefore, the purpose of proposing Example 6 is:
(1) Emphasize that only $a b=1$ is needed to derive
$$
\frac{1}{a+1}+\frac{1}{b+1}=1 .
$$

However, this is somewhat "shallow and old";
(2) Originating from the "pairing" solution method of this problem.
Solution 2: The given conditions indicate that $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so $a b=1$.
Let $M=\frac{1}{a+1}+\frac{1}{b+1}$, and pair
$$
\begin{array}{l}
N=\frac{a}{a+1}+\frac{b}{b+1} . \\
\text { Then } M+N=\frac{a+1}{a+1}+\frac{b+1}{b+1}=2, \\
M-N=\frac{1-a}{a+1}+\frac{1-b}{b+1} \\
=\frac{(1-a)(1+b)+(1+a)(1-b)}{(a+1)(b+1)} \\
=\frac{2(1-a b)}{(a+1)(b+1)}=0 .
\end{array}
$$

Then $M+N=\frac{a+1}{a+1}+\frac{b+1}{b+1}=2$,

so $M=N=1$.
By incorporating the idea of pairing, the original form of the problem is hidden (only retaining $a b=1$), thus forming the test question.
3.3 Variations

After the test question was released, literature [1], [3-5] discussed its variations, generalizations, and applications.
(1) Variations
Let $a=\frac{x}{m}, b=\frac{y}{m}$, then we have the following Example 8.