Answer :
Let the width be $w$ and the length be $l$. According to the problem, the length is three times the width. Therefore, we have
$$
l = 3w.
$$
The perimeter $P$ of a rectangle is given by
$$
P = 2w + 2l.
$$
Substitute $l = 3w$ into the formula:
$$
P = 2w + 2(3w).
$$
The expression $2w + 2(3w)$ simplifies to
$$
2w + 6w = 8w.
$$
The problem states that the perimeter is at most 112 cm, which means
$$
P \leq 112.
$$
Since $P = 8w$, we can write the inequality as
$$
8w \leq 112.
$$
In the format provided in the answer choices, this inequality corresponds to
$$
2w + 2(3w) \leq 112.
$$
Therefore, the correct inequality is
$$
\boxed{2w + 2(3w) \le 112.}
$$
$$
l = 3w.
$$
The perimeter $P$ of a rectangle is given by
$$
P = 2w + 2l.
$$
Substitute $l = 3w$ into the formula:
$$
P = 2w + 2(3w).
$$
The expression $2w + 2(3w)$ simplifies to
$$
2w + 6w = 8w.
$$
The problem states that the perimeter is at most 112 cm, which means
$$
P \leq 112.
$$
Since $P = 8w$, we can write the inequality as
$$
8w \leq 112.
$$
In the format provided in the answer choices, this inequality corresponds to
$$
2w + 2(3w) \leq 112.
$$
Therefore, the correct inequality is
$$
\boxed{2w + 2(3w) \le 112.}
$$