B2 There is a unique triplet of positive integers $(a, b, c)$ such that $a \leq b \leq c$ and
$$
\frac{25}{84}=\frac{1}{a}+\frac{1}{a b}+\frac{1}{a b c} .
$$

Determine $a+b+c$.
Solution:
Note that $\frac{1}{4}<\frac{25}{84}<\frac{1}{3}$. Therefore, $a \geq 4$. But if $a \geq 5$, then $b, c \geq 5$. Consequently,
$$
\frac{25}{84}=\frac{1}{a}+\frac{1}{a b}+\frac{1}{a b c} \leq \frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}=\frac{5^{2}+5+1}{5^{3}}=\frac{31}{125}<\frac{1}{4}<\frac{25}{84},
$$
which is a contradiction. Therefore, $a \geq 5$. Hence, $a=4$.

Substituting this into the equation given in the problem yields
$$
\frac{25}{84}=\frac{1}{4}+\frac{1}{4 b}+\frac{1}{4 b c} .
$$

Multiplying both sides by 4 and rearranging yields
$$
\frac{4}{21}=\frac{1}{b}+\frac{1}{b c}
$$

Note that $\frac{1}{6}<\frac{4}{21}<\frac{1}{5}$. Therefore, $b \geq 6$. If $b \geq 7$, then $c \geq 7$. Hence,
$$
\frac{4}{21}=\frac{1}{b}+\frac{1}{b c} \leq \frac{1}{7}+\frac{1}{7^{2}}=\frac{7+1}{7^{2}}=\frac{8}{49}<\frac{1}{6}<\frac{4}{21},
$$
which is a contradiction. Therefore, $b \nsupseteq 7$. Consequently, $b=6$. Substituting this into (1) yields
$$
\frac{4}{21}=\frac{1}{6}+\frac{1}{6 c}
$$

Multiplying both sides by 6 and rearranging yields
$$
\frac{1}{7}=\frac{1}{c}
$$

Therefore, $c=7$.

Hence, $(a, b, c)=(4,6,7)$, which yields $a+b+c=17$.