Three, (50 points) Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
P=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}, \\
Q=\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{z+x}, \\
R=\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x} .
\end{array}
$$

Let $f=\max (P, Q, R)$, find $f_{\text {min }}$.
$$
\begin{array}{l}
\text { 3, } Q-R \\
=x\left(\frac{1}{z+x}-\frac{1}{y+z}\right)+y\left(\frac{1}{x+y}-\frac{1}{z+1}\right)+\left(\frac{1}{y+z}-\frac{1}{x+y}\right) \\
=\frac{x(y-x)}{(z+x)(y+z)}+\frac{y(z-y)}{(x+y)(z+x)}+\frac{z(x-z)}{(y+z)(x+y)} \\
=\frac{x\left(y^{2}-x^{2}\right)+y\left(z^{2}-y^{2}\right)+z\left(x^{2}-z^{2}\right)}{(x+y)(y+z)(z+x)} \\
=\frac{x\left(y^{2}-x^{2}\right)+y\left(z^{2}-y^{2}\right)+z\left(x^{2}-y^{2}+y^{2}-z^{2}\right)}{(x+y)(y+z)(z+x)} \\
=\frac{\left(x^{2}-y^{2}\right)(z-x)-(y+z)(y-z)^{2}}{(x+y)(y+z)(z+x)} \\
=\frac{\left(y^{2}-z^{2}\right)(x-y)-(x+z)(x-z)^{2}}{(x+y)(y+z)(z+x)} \\
=\frac{\left(z^{2}-x^{2}\right)(y-z)-(x+y)(x-y)^{2}}{(x+y)(y+z)(z+x)} .
\end{array}
$$

For $x, y, z$ variables,
if $x$ is the largest, from equation (1) we know $Q \leqslant R$;
if $y$ is the largest, from equation (2) we know $Q \leqslant R$;
if $z$ is the largest, from equation (3) we know $Q \leqslant R$.
Therefore, for any $x, y, z \in \mathbf{R}_{+}$, we have $Q \leqslant R$.
If we swap $x$ and $z$, then $Q$ becomes $P$, and $R$ remains unchanged, so we still have $P \leqslant R$.
Thus, $f=\max (P, Q, R)=R$.
Notice that
$$
\begin{aligned}
R & =\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x} \\
& =(x+y+z)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3 \\
= & \frac{1}{2}[(x+y)+(y+z)+(z+x)] . \\
& \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3 \\
\geqslant & \frac{9}{2}-3=\frac{3}{2} .
\end{aligned}
$$

The equality holds if and only if $x=y=z$, hence
$$
f_{\text {min }}=\frac{3}{2} \text {. }
$$
（Ju Guoqi, Qianwei County No. 2 Middle School, Anhui Province, 246300）
