Ta có $P-2-\sqrt{\frac 12}= \left( \sqrt{\frac{x^2+z^2}{y^2+z^2}}+\sqrt{\frac{y^2+z^2}{x^2+z^2}}-2 \right)+ \left(\sqrt{\frac{z^2+xy}{x^2+y^2}}-\sqrt{\frac12} \right)$
$=\frac{\big(\sqrt{x^2+z^2}-\sqrt{y^2+z^2}\big)^2}{\sqrt{(x^2+z^2)(y^2+z^2)}}+\frac{2\sqrt{z^2+xy}-\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}$
$=\tfrac{(x^2-y^2)^2}{\big(\sqrt{x^2+z^2}+\sqrt{y^2+z^2} \big)^2\sqrt{(x^2+z^2)(y^2+z^2)}}+\tfrac{2z^2-(x-y)^2}{2\big(2\sqrt{z^2+xy}+\sqrt{x^2+y^2} \big)\sqrt{x^2+y^2}}$
$=(x-y)^2\Bigg(\tfrac{(x+y)^2}{\big(\sqrt{x^2+z^2}+\sqrt{y^2+z^2} \big)^2\sqrt{(x^2+z^2)(y^2+z^2)}}-\tfrac{1}{2\big(2\sqrt{z^2+xy}+\sqrt{x^2+y^2} \big)\sqrt{x^2+y^2}} \Bigg)+\tfrac{z^2}{\big(2\sqrt{z^2+xy}+\sqrt{x^2+y^2} \big)\sqrt{x^2+y^2}}$
Dễ dàng cm biểu thức trong ngoặc lớn $>0$ do $x \ge y \ge z$
Từ đó suy ra $\min P=2+\sqrt{\frac{1}{2}}\Leftrightarrow x=y,z=0$