China Southeastern Mathematical Olympiad

First, prove that $f\le \frac{1}{7}$; when $x=y=z=\frac{1}{3}$, we have $f=\frac{1}{7}$. Since $f=\sum \frac{x(x+3y-1)}{1+x+3y}=1-2\sum \frac{x}{1+x+3y}$, by Cauchy's inequality $$\sum \frac{x}{1+x+3y}\ge \frac{{\left(\sum x\right)}^2}{\sum x(1+x+3y)}=\frac{1}{\sum x(1+x+3y)},$$ and $$\sum x(1+x+3y)-\sum x(2x+4y+z)=2+\sum xy\le \frac{7}{3}.$$ So $\sum \frac{x}{1+x+3y}\ge \frac{3}{7}$, $f\le 1-2\times \frac{3}{7}=\frac{1}{7}$; $f_{max}=\frac{1}{7}$; when $x=y=z=\frac{1}{3}$, we have $f=\frac{1}{7}$.

Second, prove that $f\ge 0$; when $x=1,y=z=0$, we have $f=0$. In fact, one can see that $$\begin{array}{rl}f(x,y,z)& =\frac{x(2y-z)}{1+x+3y}+\frac{y(2z-x)}{1+y+3z}+\frac{z(2x-y)}{1+z+3x}\\ & =xy\left(\frac{2}{1+x+3y}-\frac{1}{1+y+3z}\right)\\ & +xz\left(\frac{2}{1+z+3x}-\frac{1}{1+x+3y}\right)\\ & +yz\left(\frac{2}{1+y+3z}-\frac{1}{1+z+3x}\right)\\ & =\frac{7xyz}{(1+x+3y)(1+y+3z)}\\ & +\frac{7xyz}{(1+z+3x)(1+x+3y)}\\ & +\frac{7xyz}{(1+y+3z)(1+z+3x)}\\ & \ge 0.\end{array}$$