Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which the arable land can decrease on average each year.
Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \times(1+10 \%) .
\end{array}
$$

Simplifying, we get
$$
\begin{aligned}
x \leqslant & 10^{3} \times\left[1-\frac{1.1 \times(1+0.01)^{10}}{1.22}\right] \\
= & 10^{3} \times\left[1-\frac{1.1}{1.22}\left(1+\mathrm{C}_{10}^{1} \times 0.01\right.\right. \\
& \left.\left.+\mathrm{C}_{10}^{2} \times 0.01^{2}+\cdots\right)\right] \\
\approx & 10^{3} \times\left[1-\frac{1.1}{1.22} \times 1.1045\right] \approx 4.1 .
\end{aligned}
$$

Therefore, the maximum average annual reduction of arable land in hectares is 4.