Five. (15 points) Given a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have
$$
\begin{array}{l}
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4} . \\
\text { Find }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)]
\end{array}
$$

where $[a]$ denotes the greatest integer not exceeding the real number $a$.
$$
f(1)=\frac{1}{2} \text {. }
$$

Let $y=z=0$. Then
$$
-\frac{1}{2} f(0)+\frac{1}{2} f(0)-f(x) f(0) \geqslant \frac{1}{4} \text {. }
$$

Substituting $f(0)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \leqslant \frac{1}{2} \text {. }
$$

Now let $y=z=1$. Then
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1) \geqslant \frac{1}{4} \text {. }
$$

Substituting $f(1)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \geqslant \frac{1}{2} \text {. }
$$

Combining (1) and (2), for any real number $x$,
$$
f(x)=\frac{1}{2} \text {. }
$$

Upon verification, $f(x)=\frac{1}{2}$ satisfies the given conditions.
$$
\begin{array}{l}
\text { Therefore, }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)] \\
=\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\cdots+\left[\frac{2011}{2}\right] \\
=0+1+1+\cdots+1005+1005 \\
=2(1+2+\cdots+1005) \\
=(1+1005) \times 1005 \\
=1011030 .
\end{array}
$$

Five, let $x=y=z=0$. Then
$$
\begin{array}{l}
\frac{1}{2} f(0)+\frac{1}{2} f(0)-f^{2}(0) \geqslant \frac{1}{4} \\
\Rightarrow\left(f(0)-\frac{1}{2}\right)^{2} \leqslant 0 \\
\Rightarrow f(0)=\frac{1}{2} .
\end{array}
$$

Now let $x=y=z=1$. Then similarly we have