Ta có:$\dfrac{x^2}{x+y^2}=x-\dfrac{xy^2}{x+y^2}\ge x-\dfrac{xy^2}{2\sqrt{xy^2}}=x-\dfrac{y\sqrt x}{2}\ge x-\dfrac{1}{4}(y+xy)$
Tương tự: $\dfrac{y^2}{y+z^2}\ge y-\dfrac{1}{4}(z+yz);\dfrac{z^2}{z+x^2}\ge z-\dfrac{1}{4}(x+xz)$.
Suy ra:
$P\ge\dfrac{3}{4}(x+y+z)-\dfrac{1}{4}(xy+yz+zx)\ge\dfrac{9}{4}-\dfrac{(x+y+z)^2}{12}=\dfrac{3}{2}$.
$\min P=\dfrac{3}{2} \Leftrightarrow x=y=z=1$.