For a triangle, we know that $a b + b c + c a = 12$ (where $a, b$, and $c$ are the sides of the triangle.) What are the limits within which the perimeter of the triangle can fall?
I. solution. Let $k$ be the perimeter of the triangle. Then

\[
\begin{aligned}
& k^{2}=(a+b+c)^{2}=\left(a^{2}+b^{2}+c^{2}\right)+2(a b+b c+a c)= \\
& =\frac{1}{2}(a-b)^{2}+\frac{1}{2}(b-c)^{2}+\frac{1}{2}(c-a)^{2}+3(a b+a c+b c)
\end{aligned}
\]

Considering the condition $a b+a c+b c=12$, we get

\[
2 k^{2}-72=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}
\]

The right-hand side is non-negative, so $k^{2} \geqq 36, k \geqq 6$, and equality holds only if $a=b=c$. On the other hand, if $a=b=c=2$, then $k=6$ and $a b+b c+c a=12$, so the minimum perimeter is 6.

To determine the maximum, we will also use (1). The triangle inequality can be written in the following form:

\[
|a-b| \leqq c, \quad|b-c| \leqq a \quad \text { and } \quad|a-c| \leqq b
\]

Thus, the right-hand side of (1) is not greater than $c^{2}+a^{2}+b^{2}=k^{2}-2(a b+b c+c a)=k^{2}-24$, so

\[
2 k^{2}-72 \leqq k^{2}-24, \quad k^{2} \leqq 48, \quad k \leqq \sqrt{48}=4 \sqrt{3}
\]

This gives us an upper bound for the perimeter. The question is: does $k$ take this value? Clearly, it does exactly when equality holds everywhere in (2), which requires allowing degenerate triangles. If we allow this and assume $a \geqq b \geqq c$, then we get that $k$ takes the maximum value of $4 \sqrt{3}$ exactly when $a-b=c$, $b-c=a$, and $a-c=b$. Adding the first two equations gives $c=0$, and from this (in all three equations) $a=b$ follows. Finally, the condition $a b+a c+b c=12$ with $c=0$ and $a=b$ gives $a=b=2 \sqrt{3}$.

Thus, we have found that the perimeter is at most $4 \sqrt{3}$, and it is exactly $4 \sqrt{3}$ in the degenerate isosceles triangle where the base is 0 and the two legs are $2 \sqrt{3}$. The perimeter value therefore varies between 6 and $4 \sqrt{3}$.

II. solution. According to the relationship between the arithmetic and quadratic means,

\[
\left(\frac{k}{3}\right)^{2}=\left(\frac{a+b+c}{3}\right)^{2} \leqq \frac{a^{2}+b^{2}+c^{2}}{3}
\]

so

\[
k^{2} \leqq 3\left(a^{2}+b^{2}+c^{2}\right)=3\left(k^{2}-2(a b+b c+c a)\right)=3 k^{2}-72
\]

and from this $k^{2} \geqq 36, k \geqq 6$. Here, equality holds if and only if $a=b=c=2$, so the minimum perimeter is 6.

To find the maximum, assume $a \geqq b \geqq c$. Then, by the triangle inequality, $a \leqq b+c$, and multiplying by the non-negative $a$ gives

\[
a^{2} \leqq a b+a c \leqq a b+a c+b c=12
\]

Furthermore, since $b \geqq c \geqq 0$, $c^{2} \leqq b c$, and since $a \geqq b \geqq 0$, $b^{2} \leqq a b$, so

\[
b^{2}+c^{2} \leqq a b+b c \leqq a b+a c+b c=12
\]

Adding the two inequalities,

\[
a^{2}+b^{2}+c^{2} \leqq 24
\]

or $k^{2}-2(a b+a c+b c) \leqq 24, k^{2}-24 \leqq 24, k^{2} \leqq 48$. If we check under what conditions equality holds in the added two inequalities, we can easily see that it is only when $c=0, b=a$.

Remark. It can be shown that the perimeter takes any value between 6 and $4 \sqrt{3}$, and except for $4 \sqrt{3}$, every value is taken in a non-degenerate triangle.