Đặt $a=\dfrac{y}{z} \ \ \ \ \ b=\dfrac{z}{x} \rightarrow c=\dfrac{x}{y}$
$\rightarrow$ $\dfrac{ab}{2b+c}=\dfrac{y^2}{2zy+x^2} \\ \dfrac{bc}
{2c+a}=\dfrac{z^2}{2xz+y^2} \\ \dfrac{ca}{2a+b}=\dfrac{x^2}{2xy+z^2}$
$\rightarrow P=\dfrac{y^2}{2zy+x^2}+\dfrac{z^2}{2xz+y^2}+
\dfrac{x^2}{2xy+z^2}$
Áp dụng bất đẳng thức $2mn \leq m^2+n^2$ ta có:
$P=\dfrac{y^2}{2zy+x^2}+\dfrac{z^2}{2xz+y^2}+ \dfrac { x^2 } { 2xy+z^2 }
\geq \dfrac{y^2}{z^2+y^2+x^2}+\dfrac{z^2}{x^2+z^2+y^2}+\dfrac{x^2}{x^2+y^2+z^2}=1$
Đẳng thức xảy ra $\leftrightarrow x=y=z \leftrightarrow a=b=c=1$
Vậy $MinP = 1 \leftrightarrow a=b=c=1$