10) (20 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of $f(x, y, z) = (1-yz+z)(1-zx+x)(1-xy+y)$ and the values of $x, y, z$ at that time.
10 Notice
$$
\left\{\begin{array} { l } 
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right.
$$
$\Rightarrow(x+x z)(y+x y)<1 \Leftrightarrow x+x y+x^{2} y<0$, contradiction.
Therefore, $1-y z+z$, $1-z x+x$, and $1-x y+y$ can have at most one negative value.
Assume all three expressions are positive. Then,
$$
\begin{aligned}
& (1-y z+z)(1-z x+x)(1-x y+y) \\
= & (x-x y z+x z)(y-x y z+x y)(z-x y z+y z) \\
= & (x-1+x z)(y-1+x y)(z-1+y z),
\end{aligned}
$$

Thus,
$$
\begin{aligned}
& f^{2}(x, y, z) \\
= & {[(1-y z+z)(1-x z+x)(1-x y+y)]^{2} } \\
= & {[(1-y z+z)(1-x z+x)(1-x y+y)] . } \\
& {[(x-1+x z)(y-1+x y)(z-1+y z)] } \\
= & {\left[z^{2}-(1-y z)^{2}\right]\left[x^{2}-(1-x z)^{2}\right]\left[y^{2}-(1-x y)^{2}\right] } \\
\leqslant & (x y z)^{2}=1,
\end{aligned}
$$

When $x=y=z=1$, equality holds.
Therefore, $f(x, y, z)_{\text {max }}=1$.
11 (1) Let $P(a \cos \alpha, b \sin \alpha)$, $Q(a \cos \beta, b \sin \beta)$, and by $A(-a, 0)$, $B(a, 0)$ we get
$$
\begin{array}{l}
l_{A P}: a(1+\cos \alpha) y=b \sin \alpha(x+a), \\
l_{Q B}: a(\cos \beta-1) y=b \sin \beta(x-a),
\end{array}
$$

Solving (1) and (2) to eliminate $y$ gives
$$
\begin{aligned}
& \sin \alpha(\cos \beta-1)(x+a) \\
= & \sin \beta(1+\cos \alpha)(x-a) \\
\Leftrightarrow & {[\sin \alpha(\cos \beta-1)-\sin \beta(1+\cos \alpha)] x } \\
= & a[\sin \alpha(1-\cos \beta)-\sin \beta(1+\cos \alpha)] \\
\Leftrightarrow & {[\sin (\alpha-\beta)-\sin \alpha-\sin \beta] x } \\
= & a[\sin \alpha-\sin \beta-\sin (\beta+\alpha)] \\
\Leftrightarrow & \cos \frac{\alpha-\beta}{2}\left(\sin \frac{\alpha-\beta}{2}-\sin \frac{\alpha+\beta}{2}\right) x \\
= & a \cos \frac{\alpha+\beta}{2}\left(\sin \frac{\alpha-\beta}{2}-\sin \frac{\alpha+\beta}{2}\right) \\
\Leftrightarrow & x_{M}=\frac{a \cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}} \text { (since } P 、 Q \text { are not the vertices). }
\end{aligned}
$$

Similarly, $x_{N}=\frac{a \cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}}$, hence $x_{M}=x_{N}$, so $M N$
$\perp A B$,