Ta có:
$P+3=\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}+\dfrac{x+y+z}{x+y}$
$=(x+y+z)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)$
$\ge(x+y+z).\dfrac{9}{2(x+y+z)}=\dfrac{9}{2}$
$\Rightarrow P\ge\dfrac{3}{2}$
$\min P=\dfrac{3}{2} \Leftrightarrow x=y=z$