3. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have $f(x y)+f(x z)-2 f(x) f(y z) \geqslant \frac{1}{2}$, then $[1 \cdot f(1)]+[2 \cdot f(2)]+\cdots+[2022 \cdot f(2022)]=$ $\qquad$ . (where $[x]$ denotes the greatest integer less than or equal to $x$)
Let $x=y=z=1 \Rightarrow 2 f(1)-2 f^{2}(1) \geqslant \frac{1}{2} \Rightarrow (2 f(1)-1)^{2} \leqslant 0 \Rightarrow f(1)=\frac{1}{2}$, similarly we get $f(0)=\frac{1}{2}$. Let $y=z=1 \Rightarrow 2 f(x)-2 f(x) f(1) \geqslant \frac{1}{2} \Rightarrow f(x) \geqslant \frac{1}{2}$, let $y=z=0 \Rightarrow 2 f(0)-2 f(x) f(0) \geqslant \frac{1}{2} \Rightarrow f(x) \leqslant \frac{1}{2}$, thus $f(x)=\frac{1}{2}$. Therefore, $[1 \cdot f(1)]+[2 \cdot f(2)]+\cdots+[2022 \cdot f(2022)]$ $=2(1+2+\cdots+1010)+1011=1011^{2}=1022121$.