Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow xyz=1\)
\(P=\frac{x^2}{z}+\frac{y^z}{x}+\frac{z^2}{y}+\frac{9}{2\left(xy+xz+yz\right)}\) (do \(\frac{9}{2\left(a+b+c\right)}=\frac{9}{\frac{2\left(a+b+c\right)}{abc}}=...\))
\(\frac{x^2}{z}+z\ge2x\); \(\frac{y^2}{x}+x\ge2y\); \(\frac{z^2}{y}+y\ge2z\)
\(\Rightarrow\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\ge x+y+z\)
\(\Rightarrow P\ge x+y+z+\frac{9}{2\left(xy+xz+yz\right)}\ge x+y+z+\frac{27}{2\left(x+y+z\right)^2}\)
\(P\ge\frac{x+y+z}{2}+\frac{x+y+z}{2}+\frac{27}{2\left(x+y+z\right)^2}\ge3\sqrt[3]{\frac{27\left(x+y+z\right)^2}{8\left(x+y+z\right)^2}}=\frac{3}{2}\)
\(\Rightarrow P_{min}=\frac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)