21. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$, we have $\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant$
$$
\frac{1}{4} \text {. }
$$
21. The given is an inequality, not an equation, and there are three variables, namely $x, y, z$. We attempt to find the result by taking some special values.
Let $x=y=z=1$, substitute into (1), we get
$$
\begin{array}{c}
f(1)-[f(1)]^{2} \geqslant \frac{1}{4} . \\
{\left[f(1)-\frac{1}{2}\right]^{2} \leqslant 0 .} \\
f(1)=\frac{1}{2} .
\end{array}
$$

So
Let $y=z=1$, substitute into (1) and use (2), we get
$$
f(x)-\frac{1}{2} f(x) \geqslant \frac{1}{4} .
$$

So
$$
f(x) \geqslant \frac{1}{2} .
$$

Let $x=y=z=0$, substitute into (1), we get
$$
f(0)-f^{2}(0) \geqslant \frac{1}{4} \text {. }
$$

So
$$
f(0)=\frac{1}{2} \text {. }
$$

Let $x=0$, substitute into (1) and use (4), we get
$$
\frac{1}{2}-\frac{1}{2} f(y z) \geqslant \frac{1}{4} \text {. }
$$

Hence
$$
\begin{array}{l}
f(y z) \leqslant \frac{1}{2}, \\
f(x) \leqslant \frac{1}{2} .
\end{array}
$$

Combining (3) and (5), we get $f(x) \equiv \frac{1}{2}$.