Example 7 Find the minimum value of the ratio of a three-digit number (in decimal notation) to the sum of its digits.



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Let the three-digit number be $100x + 10y + z$, where $x, y, z$ are integers, and $1 \leqslant x \leqslant 9, 0 \leqslant y, z \leqslant 9$. Thus, the ratio of the number to the sum of its digits is
$$
f(x, y, z) = \frac{100x + 10y + z}{x + y + z} = 1 + \frac{99x + 9y}{x + y + z}.
$$

Since the expression on the right-hand side of (1) contains only $z$ in the denominator, $f(x, y, z)$ achieves its minimum value when and only when $z = 9$, given that $x, y$ are fixed. Similarly, we have
$$
f(x, y, z) \geqslant 1 + \frac{99x + 9y}{x + y + z} = 10 + \frac{90x - 81}{9 + x + y} \geqslant 10 + \frac{90x - 81}{18 + x} = 100 - \frac{1701}{18 + x},
$$

where the equality in the second inequality holds if and only if $y = 9$. Finally, from (2), it is clear that $f(x, y, z)$ achieves its minimum value $f(1, 9, 9) = \frac{199}{19}$ when and only when $x = 1, y = 9, z = 9$. Therefore, the minimum value sought is $\frac{199}{19}$. Note that the method used in this example is also known as the method of successive extremization.