Example 3 Let $x, y, z>0$ and $4x+3y+5z=1$, find the minimum value of $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}$, and indicate the corresponding values of $x, y, z$. (Mathematical Problems 1625, Mathematics Bulletin)
Solution: $\because 4 x+3 y+5 z=(x+y)+2(y+$ $z)+3(z+x)=1$, by Cauchy-Schwarz inequality, we get $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=(4 x+3 y+$
$5z) \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=[(x+y)+2$ $(y+z)+3(z+x)]$. $\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geqslant(1+\sqrt{2}+\sqrt{3})^{2}$, equality holds if and only if $(x+y)^{2}=2(y+z)^{2}=3(z$ $+x)^{2}$ and $4 x+3 y+5 z=1$, yielding $x=$ $\frac{6+2 \sqrt{3}-3 \sqrt{2}}{12(1+\sqrt{2}+\sqrt{3})}, y=\frac{6+3 \sqrt{2}-2 \sqrt{3}}{12(1+\sqrt{2}+\sqrt{3})}, \quad z=$ $\frac{2 \sqrt{3}+3 \sqrt{2}-6}{12(1+\sqrt{2}+\sqrt{3})}$. Therefore, the minimum value of $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}$ is $(1+\sqrt{2}+\sqrt{3})^{2}$.