China Girls' Mathematical Olympiad

The answer is $12\sqrt{2}-17$. Set $$\{\begin{array}{l}x=a+2b+c,\\ y=a+b+2c,\\ z=a+b+3c.\end{array}$$

$$\begin{array}{rl}& \frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}\\ & =\frac{2y-x}{x}+\frac{4(x+z-2y)}{y}-\frac{8(z-y)}{z}\\ & =-17+2\frac{y}{x}+4\frac{x}{y}+4\frac{z}{y}+8\frac{y}{z}\\ & \ge -17+2\sqrt{8}+2\sqrt{32}=-17+12\sqrt{2}.\end{array}$$ The equality holds if and only if $\frac{2y}{x}=\frac{4x}{y}$ and $\frac{4z}{y}=\frac{8y}{z}$, or $4x^2=2y^2=z^2$. Hence the equality holds if and only if $$\{\begin{array}{l}a+b+2c=\sqrt{2}(a+2b+c),\\ a+b+3c=2(a+2b+c).\end{array}$$ Solving the above system of equations for $b$ and $c$ in terms of $a$ gives $$\{\begin{array}{l}b=(1+\sqrt{2})a,\\ c=(4+3\sqrt{2})a.\end{array}$$ We conclude that $$\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}$$ has minimum value $12\sqrt{2}-17$ if and only if $(a,b,c)=(a,(1+\sqrt{2})a,(4+3\sqrt{2})a)$.