Ta có$\left.\begin{matrix} (x-1)^2 \ge \frac{6x-7}{9} \\ (y-2)^2 \ge \frac{12y-28}{9} \\ (z-1)^2 \ge \frac{6z-7}{9} \\ \end{matrix}\right\}\Rightarrow (x-1)^2+(y-2)^2+(z-1)^2 \ge \tfrac{2(x+y+2)+2(z+y+2)-22}{3}$
Dùng bdt $\sqrt{a^3+1}=\sqrt{(a+1)(a^2-a+1)} \le \frac{a^2+2}{2}$
Suy ra $\frac{12}{\sqrt{(\sqrt{x+y})^3+1}} \ge \frac{24}{x+y+2}$
Tương tự cho cái kia rồi ghép cặp cosi là xong
$\min P=\frac{26}3\Leftrightarrow \begin{cases}x=z=\frac 43 \\ y=\frac 83 \end{cases}$