10. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, if for any real numbers $x, y, z$, we have $\frac{1}{3} f(x y)+\frac{1}{3} f(x z)-f(x) f(y z) \geqslant \frac{1}{9}$, find $\sum_{i=1}^{100}[i f(i)]$. Here $[x]$ represents the greatest integer not exceeding $x$.
Solve:
$$
3 f(x y)+3 f(x z)-9 f(x) f(y z) \geqslant 1 \text {, }
$$

Take $x=y=z=1 \Rightarrow 9 f^{2}(1)-6 f(1)+1 \leqslant 0 \Rightarrow f(1)=\frac{1}{3}$,
Take $x=y=z=0 \Rightarrow 9 f^{2}(0)-6 f(0)+1 \leqslant 0 \Rightarrow f(0)=\frac{1}{3}$,
Take $y=z=1 \Rightarrow 6 f(x)-3 f(x) \geqslant 1 \Rightarrow f(x) \geqslant \frac{1}{3}$,
Take $y=z=0 \Rightarrow 2-3 f(x) \geqslant 1 \Rightarrow f(x) \leqslant \frac{1}{3}$, then $f(x)=\frac{1}{3}$.
Thus, $\sum_{i=1}^{100}[i f(i)]=3(1+2+\cdots+32)+2 \cdot 33=3 \cdot \frac{33 \cdot 32}{2}+2 \cdot 33=1650$.