37 Let $a$, $b$, $c$ all be positive integers greater than 1. Find the minimum possible value of the algebraic expression $\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c}$.
37. By symmetry, we may assume $a \geqslant b \geqslant c$. Note that when $(a, b, c)=(2,2,2),(3,2,2),(3,3,2),(4,2,2)$, the value of the given algebraic expression $A$ is $2, \frac{3}{2}, \frac{17}{8}, \frac{11}{4}$, respectively. This indicates that when $a+b+c \leqslant 8$, $A \geqslant \frac{3}{2}$.

Next, we prove that when $a+b+c \geqslant 9$, we have $A \geqslant \frac{3}{2}$.
In fact,
$$\begin{aligned}
A \geqslant \frac{3}{2} & \Leftrightarrow (a+b+c)^{2}-2([a, b]+[b, c]+[c, a]) \geqslant 3(a+b+c) \\
& \Leftrightarrow a^{2}+b^{2}+c^{2}+2 \sum(a b-[a, b]) \geqslant 3(a+b+c)
\end{aligned}$$

Since for positive integers $x, y$, we have $x y \geqslant [x, y]$, it suffices to prove:
$$a^{2}+b^{2}+c^{2} \geqslant 3(a+b+c)$$

Given $a+b+c \geqslant 9$, to prove (1) holds, it suffices to prove:
$$\begin{aligned}
& a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}(a+b+c)^{2} \\
\Leftrightarrow & 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant (a+b+c)^{2} \\
\Leftrightarrow & 2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a) \geqslant 0 \\
\Leftrightarrow & (a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0
\end{aligned}$$

The last inequality is obviously true.
Therefore, the minimum value of the given algebraic expression is $\frac{3}{2}$.