10. Given that $x$, $y$, and $z$ are all real numbers greater than 0 and less than 1. Then the value of $x(1-y)+y(1-z)+z(1-x)$ is ( ).
(A) greater than 1
(B) equal to 1
(C) less than 1
(D) greater than or equal to 1
10. C.

Given an equilateral triangle $\triangle A B C$ with side length 1 (as shown in Figure 7), $D, E, F$ are points on sides $A C, A B, B C$ respectively.
$$
\begin{array}{l}
\text { Let } B F=x, A E=y, \\
C D=z .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } S_{\triangle B E F}=\frac{\sqrt{3}}{4} x(1-y), \\
S_{\triangle O E 1}=\frac{\sqrt{3}}{4} y(1-z), \\
S_{\triangle O F C}=\frac{\sqrt{3}}{4} z(1-x) .
\end{array}
$$

Therefore, $S_{\triangle B E F}+S_{\triangle D E C}+S_{\triangle D F C}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{4}[x(1-y)+y(1-z)+z(1-x)] \\
<S_{\triangle \triangle B C}=\frac{\sqrt{3}}{4} .
\end{array}
$$

Thus, $x(1-y)+y(1-z)+z(1-x)<1$.