17. Given positive numbers $a, b, c$ satisfying $2a + 4b + 7c \leqslant 2abc$.
Find the minimum value of $a + b + c$.
17. When $a=3, b=\frac{5}{2}, c=2$, $a+b+c$ takes the minimum value $\frac{15}{2}$.

When $a=3, b=\frac{5}{2}, c=2$, $a+b+c=3+$ $\frac{5}{2}+2=\frac{15}{2}$.
The following is the proof: $\frac{15}{2}$ is the required minimum value.
Let $a=3 x, b=\frac{5}{2} y, c=2 z$.
From the given, we have
$$
\begin{array}{l}
2 \times 3 x+4 \times \frac{5}{2} y+7 \times 2 z \\
\leqslant 2 \times 3 x \times \frac{5}{2} y \times 2 z,
\end{array}
$$

which is $3 x+5 y+7 z \leqslant 15 x y z$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
3 x+5 y+7 z \\
=\underbrace{x+\cdots+x}_{3 \uparrow}+\underbrace{y+\cdots+y}_{5 \uparrow}+\underbrace{z+\cdots+z}_{7 \uparrow} \\
\geqslant 15 \sqrt[15]{x^{3} y^{5} z^{7}} .
\end{array}
$$

Therefore, $15 \sqrt[15]{x^{3} y^{5} z^{7}} \leqslant 15 x y z$.
Thus, $x^{6} y^{5} z^{4} \geqslant 1$.
Hence, $a+b+c=3 x+\frac{5}{2} y+2 z$
$$
\begin{array}{l}
=\frac{1}{2}(\underbrace{x+\cdots+x}_{6 \uparrow}+\underbrace{y+\cdots+y}_{5 \uparrow}+\underbrace{z+\cdots+z}_{4 \uparrow}) \\
\geqslant \frac{1}{2} \times 15 \sqrt[15]{x^{6} y^{5} z^{4}} \\
\geqslant \frac{1}{2} \times 15 \sqrt[15]{1} \geqslant \frac{15}{2} .
\end{array}
$$