17 Given positive numbers $a$, $b$, $c$ satisfy: $2a + 4b + 7c \leqslant 2abc$, find the minimum value of $a + b + c$.
17 When $a=3, b=\frac{5}{2}, c=2$, we have $a+b+c=3+\frac{5}{2}+2=\frac{15}{2}$.
Below we prove that $\frac{15}{2}$ is the minimum value sought.
Let $a=3 x, b=\frac{5}{2} y, c=2 z$, then by the given condition we have
$$
2 \cdot 3 x+4 \cdot \frac{5}{2} y+7 \cdot 2 z \leqslant 2 \cdot 3 x \cdot \frac{5}{2} y \cdot 2 z,
$$

which simplifies to
$$
3 x+5 y+7 z \leqslant 15 x y z .
$$

By the AM-GM inequality, we get
$$
\begin{aligned}
3 x+5 y+7 z & =\underbrace{x+\cdots+x}_{3}+\underbrace{y+\cdots+y}_{5}+\underbrace{z+\cdots+z}_{7} \\
& \geqslant 15 \sqrt[15]{x^{3} y^{5} z^{7}},
\end{aligned}
$$

thus
$$
\begin{aligned}
15 \sqrt[15]{x^{3} y^{5} z^{7}} & \leqslant 15 x y z, \\
x^{6} y^{5} z^{4} & \geqslant 1 .
\end{aligned}
$$

Therefore
$$
\begin{aligned}
a+b+c & =3 x+\frac{5}{2} y+2 z \\
& =\frac{1}{2}(\underbrace{x+\cdots+x}_{6}+\underbrace{y+\cdots+y}_{5}+\underbrace{z+\cdots+z}_{4}) \\
& \geqslant \frac{1}{2} \cdot 15 \sqrt[15]{x^{6} y^{5} z^{4}} \\
& \geqslant \frac{1}{2} \cdot 15 \sqrt[15]{1} \\
& \geqslant \frac{15}{2} .
\end{aligned}
$$