11. (20 points) Let $x, y, z \in(0,1)$, and $x^{2}+y^{2}+z^{2}=1$.
Determine the maximum value of $f=x+y+z-x y z$.

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11. When $x=y=z=\frac{\sqrt{3}}{3}$, $f=\frac{8 \sqrt{3}}{9}$.

Next, we prove: $f \leqslant \frac{8 \sqrt{3}}{9}$.
Let $a=\sqrt{3} x, b=\sqrt{3} y, c=\sqrt{3} z$. Then $a^{2}+b^{2}+c^{2}=3$.

Hence, (1) $\Leftrightarrow 3(a+b+c)-a b c \leqslant 8$.
Notice that
$$
a b c=\frac{1}{3}\left[\sum a^{2}-\sum a\left(\sum a^{2}-\sum a b\right)\right] .
$$

Thus, (2)
$\Leftrightarrow\left(\sum a\right)\left(3+\frac{3 \sum a^{2}-\sum a b}{6}\right) \leqslant 8+\frac{\sum a^{3}}{3}$
$\Leftrightarrow(\searrow a)\left[3+\frac{3 \sum a^{2}-(\searrow a)^{2}}{6}\right] \leqslant 8+\frac{\searrow}{3}$ a
Let it be, $a=x$. Then we only need to prove
$$
x\left(\frac{9}{2}-\frac{x^{2}}{6}\right) \leqslant 8+\frac{\sum a^{3}}{3} \text {. }
$$

By the AM-GM inequality, we get
$$
\begin{array}{l}
x\left(\frac{9}{2}-\frac{x^{2}}{6}\right)=\frac{1}{2} x\left(9-\frac{x^{2}}{3}\right) \\
=\frac{1}{2} \sqrt{\frac{3}{2}} \times \frac{2 x^{2}}{3}\left(9 \quad \frac{x^{2}}{3}\right) \\
\leqslant \frac{1}{2} \sqrt{3}\binom{18}{2}-9 \text {. } \\
\end{array}
$$
$$
\left(\sum a\right)\left(\sum a^{\prime}\right) \geqslant\left(\sum a a^{2}=9\right. \text {. }
$$

If $a \leqslant \sqrt{3} \sum a=3$, then บ $a^{3} \geqslant 3$.
Thus, $8+\frac{\searrow}{3} a^{3} \geqslant 9 \geqslant x\left(\frac{9}{2}-\frac{x^{2}}{6}\right)$.
Therefore, inequality (3) is proved:
Hence, $f_{\text {max }}=\frac{8 \sqrt{3}}{9}$.