43. Let $f(x, y, z)=\frac{x(2 y-z)}{1+x+3 y}+\frac{y(2 z-x)}{1+y+3 z}+\frac{z(2 x-y)}{1+z+3 x}$, where $x, y, z \geqslant 0, x+$ $y+z=1$, find the maximum and minimum values of $f(x, y, z)$. (2009 China Southeast Mathematical Olympiad
43. Proof: $f(x, y, z) \leqslant \frac{1}{7}$. Equality holds if and only if $x=y=z=\frac{1}{3}$.

Since $\frac{x(2 y-z)}{1+x+3 y}=\frac{x(-2 y-(1-x-y))}{1+x+3 y}=\frac{x(-1+x+3 y)}{1+x+3 y}=x- \frac{2 x}{1+x+3 y}$, we have
$$\begin{aligned}
f(x, y, z)= & (x+y+z)-2\left(\frac{x}{1+x+3 y}+\frac{y}{1+y+3 z}+\frac{z}{1+z+3 x}\right)= \\
& 1-2\left(\frac{x}{1+x+3 y}+\frac{y}{1+y+3 z}+\frac{z}{1+z+3 x}\right)= \\
& 1-2 \sum \frac{x}{1+x+3 y}
\end{aligned}$$

By the Cauchy-Schwarz inequality, we get $\sum \frac{x}{1+x+3 y} \geqslant \frac{\left(\sum x\right)^{2}}{\sum x(1+x+3 y)}=\frac{1}{\sum x(1+x+3 y)}$.
$$\begin{aligned}
\sum x(1+x+3 y)= & \sum x(2 x+3 y+z)= \\
& 2\left(\sum x\right)^{2}+\sum x y \leqslant \frac{7}{3}
\end{aligned}$$

Thus, $\sum \frac{x}{1+x+3 y} \geqslant \frac{3}{7}$, and hence $f(x, y, z) \leqslant \frac{1}{7}$.
That is, $f(x, y, z)_{\text {max }}=\frac{1}{7}$, with equality if and only if $x=y=z=\frac{1}{3}$.
Next, we prove $f(x, y, z) \geqslant 0$, with equality when $x=1, y=z=0$.
Let $z=\min \{x, y, z\}$. If $z=0$, then
$$f(x, y, 0)=\frac{2 x y}{1+x+3 y}-\frac{x y}{1+y}=\frac{2 x y}{2 x+4 y}-\frac{x y}{x+2 y}=0$$

Now assume $x, y \geqslant z>0$. By $(*)$, to prove $f(x, y, z) \geqslant 0$, it suffices to show
$$\sum \frac{x}{1+x+3 y} \leqslant \frac{1}{2}$$

Note that $\frac{1}{2}=\frac{x}{2 x+4 y}+\frac{y}{x+2 y}$, so inequality (1) is equivalent to
$$\begin{aligned}
\frac{z}{1+z+3 x} \leqslant & \frac{x}{2 x+4 y}-\frac{x}{1+x+3 y}+\frac{y}{x+2 y}-\frac{y}{1+y+3 z}= \\
& \frac{z}{2 x+4 y}\left(\frac{x}{1+x+3 y}+\frac{8 y}{1+y+3 z}\right)
\end{aligned}$$
$$\frac{2 x+4 y}{1+z+3 x} \leqslant \frac{x}{1+x+3 y}+\frac{8 y}{1+y+3 z}$$

By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
\frac{x}{1+x+3 y}+\frac{8 y}{1+y+3 z}= & \frac{x^{2}}{x(1+x+3 y)}+\frac{(2 y)^{2}}{\frac{y(1+y+3 z)}{2}} \geqslant \\
& \frac{(x+2 y)^{2}}{x(1+x+3 y)+\frac{y(1+y+3 z)}{2}}= \\
& \frac{2 x+4 y}{1+z+3 x}
\end{aligned}$$

Therefore, $f(x, y, z) \geqslant 0$. That is, the minimum value of $f(x, y, z)$ is 0. Equality holds if and only if $x=1, y=z=0$ or $y=1, z=x=0$ or $z=1, x=y=0$.