First let's try to prove that \([a, b]|c \) implies \( a|c\) and \(b|c\).
\([a,b]|c\) means that some multiple of \(a\) divides \(c\) and that some multiple of \(b\) divides \(c\):
$$ \begin{align} [a,b]|c &\implies am|c, bn|c \\ &\implies a|c, b|c \end{align} $$
Now let's try to prove the converse. We can represent the prime factorization of \(a\), \(b\) and \(c\) like this:
$$ \begin{align} a &= {p_1}^{x_1} {p_2}^{x_2} {p_3}^{x_3} \cdots \\ b &= {p_1}^{y_1} {p_2}^{y_2} {p_3}^{y_3} \cdots \\ c &= {p_1}^{z_1} {p_2}^{z_2} {p_3}^{x_3} \cdots \end{align} $$
Since \([a,b]\) is the least common multiple:
$$ [a,b] = {p_1}^{\max(x_1,y_1)} {p_2}^{\max(x_2,y_2)} {p_3}^{\max(x_3,y_3)} \cdots $$
If \(a|c\), then \(x_i \le z_i\) for all \(i \ge 1\). If \(b|c\), then \(y_i \le z_i\) for all \(i \ge 1\). If both \(a\) and \(b\) divides \(c\), then \(\max(x_i, y_i) \le z_i\) for all \(i \ge 1\). Hence, \([a,b]|c\).