$13 \cdot 17$ When $a, b$ are two unequal positive numbers, among the following three algebraic expressions:
甲: $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)$, 乙: $\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}$,
丙: $\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}$
the one with the largest value
(A) must be 甲.
(B) must be 乙.
(C) must be 丙.
(D) is generally not certain, and depends on the values of $a, b$. (China High School Mathematics League, 1982)
[Solution 1] First, compare expressions 甲 and 乙. By the problem statement,
$$
\text { 甲 }=\frac{1}{a b}\left(a^{2} b^{2}+a^{2}+b^{2}+1\right), \quad \text { 乙 }=\frac{1}{a b}\left(a^{2} b^{2}+2 a b+1\right) \text {. }
$$

Since $a^{2}+b^{2}>2 a b$, we know 甲 $>$ 乙. Next, compare 甲 and 丙. By
$$
\text { 丙-甲 }=\frac{(a-b)^{2}}{4 a b(a+b)^{2}}\left[(a+b)^{2}(a b-4)-4\right] \text {, }
$$

and since the sign of $(a+b)^{2}(a b-4)-4$ varies depending on the values of $a, b$ relative to 4, it is uncertain whether 甲 or 丙 is larger.
Therefore, the answer is $(D)$.
[Solution 2] In Solution 1, the calculation of 丙 - 甲 is relatively complex. We can estimate 甲 and 丙 using the following method:

Let $b=1$, then 甲 $=2\left(a+\frac{1}{a}\right)$, and 丙 $=\left(\frac{a+1}{2}+\frac{2}{a+1}\right)^{2}$. From this, we can see that when $a$ is very large, 甲 $\approx 2 a$, and 丙 $\approx \frac{1}{4}(a+1)^{2}>\frac{1}{4} a \cdot a>2 a$.
For example, if $a=9$, then 甲 $=25$. In this case, 丙 $>$ 甲.
If $a=\frac{1}{5}$, then by the above expressions, 甲 $>10$, and 丙 $\leqslant(1+2)^{2}=9$. In this case, 甲 $>$ 丙. Therefore, it is uncertain whether 甲 or 丙 is larger.
Hence, the answer is $(D)$.