6. Given that $a$, $b$, and $c$ are the lengths of the sides of a given triangle. If positive real numbers $x$, $y$, and $z$ satisfy $x+y+z=1$, find the maximum value of $a x y + b y z + c z x$.
6. Let's assume $00$, and
$p+q=a, q+r=b, r+p=c, q \leqslant p \leqslant r$.
Thus, $f(x, y, z)$
$=(p+q) x y+(q+r) y z+(r+p) z x$
$=p x(y+z)+q y(z+x)+r z(x+y)$
$=p x(1-x)+q y(1-y)+r z(1-z)$
$=\frac{p+q+r}{4}-$
$$
\begin{aligned}
& \left(p\left(\frac{1}{2}-x\right)^{2}+q\left(\frac{1}{2}-y\right)^{2}+r\left(\frac{1}{2}-z\right)^{2}\right) \\
= & \frac{a+b+c}{8}- \\
& \left(p\left(\frac{1}{2}-x\right)^{2}+q\left(\frac{1}{2}-y\right)^{2}+r\left(\frac{1}{2}-z\right)^{2}\right) .
\end{aligned}
$$

Let $m=\left|\frac{1}{2}-x\right|, n=\left|\frac{1}{2}-y\right|, l=\left|\frac{1}{2}-z\right|$.
Then $m, n, l \geqslant 0, m+n+l \geqslant \frac{1}{2}$.
Let $g(m, n, l)=p m^{2}+q n^{2}+r l^{2}$.
We now find the minimum value of $g(m, n, l)$.
Notice,
$$
\begin{array}{l}
\left(p m^{2}+q n^{2}+r l^{2}\right)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right) \\
\geqslant(m+n+l)^{2} \geqslant \frac{1}{4} \\
\Rightarrow p m^{2}+q n^{2}+r l^{2} \geqslant \frac{p q r}{4(p q+q r+r p)} .
\end{array}
$$

When $p m=q n=r l$, i.e.,
$$
\begin{array}{l}
m=\frac{1}{2} \cdot \frac{q r}{p q+q r+r p}, n=\frac{1}{2} \cdot \frac{p r}{p q+q r+r p}, \\
l=\frac{1}{2} \cdot \frac{p q}{p q+q r+r p}
\end{array}
$$

the equality in (2) holds.
$$
\begin{array}{l}
\text { Hence } f(x, y, z)=\frac{a+b+c}{8}-g(m, n, l) \\
\leqslant \frac{a+b+c}{8}-\frac{p q r}{4(p q+q r+r p)} . \\
\text { When } x=\frac{1}{2} \cdot \frac{p q+r p}{p q+q r+r p}, y=\frac{1}{2} \cdot \frac{p q+q r}{p q+q r+p p}, \\
z=\frac{1}{2} \cdot \frac{q r+r p}{p q+q r+p}
\end{array}
$$

the equality in (3) holds.
From the choice in (1), we have
$$
\begin{array}{l}
8 p q r=\prod(a+b-c), \\
4(p q+q r+r p) \\
=\sum(a+b-c)(a+c-b) \\
=2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2},
\end{array}
$$

where $\sum$ denotes the cyclic sum and $\Pi$ denotes the cyclic product.
$$
\begin{array}{l}
\text { Then }(a+b+c)\left(2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}\right)- \\
\quad 8 p q r \\
=2 a b c . \\
\text { Hence } f(x, y, z) \leqslant \frac{a+b+c}{8}-\frac{p q r}{4(p q+q r+r p)} \\
=\frac{a b c}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}},
\end{array}
$$

i.e., the maximum value of $a x y+b y z+c z x$ is
$a b c$
$$
\frac{a b c}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}} \text {. }
$$

[Note] Construct the Lagrange function
$Q=a x y+b y z+c z x+\lambda(x+y+z-1)$.
Taking partial derivatives, we get
$$
\left\{\begin{array}{l}
\frac{\partial Q}{\partial x}=a y+c z+\lambda=0, \\
\frac{\partial Q}{\partial y}=a x+b z+\lambda=0, \\
\frac{\partial Q}{\partial z}=b y+c x+\lambda=0, \\
x+y+z=1 .
\end{array}\right.
$$

Solving the system of equations, we get
$$
\left\{\begin{array}{l}
x=\frac{b(a+c-b)}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}}, \\
y=\frac{c(a+b-c)}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}}, \\
z=\frac{a(b+c-a)}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}} .
\end{array}\right.
$$

Since $a, b, c$ are the lengths of the sides of a triangle, $x>0, y>0, z>0$ satisfy the conditions.
Thus, $a x y+b y z+c z x$
$$
\begin{array}{l}
=\sum a \frac{b(a+c-b) c(a+b-c)}{\left(2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}\right)^{2}} \\
=\frac{a b c}{2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}},
\end{array}
$$

i.e., the maximum value of $a x y+b y z+c z x$ is
$a b c$