\(2\sqrt{ab}\le a+b\le c\Rightarrow c^2\ge4ab\Rightarrow\frac{c^2}{ab}\ge4\)
\(P=1+\left(\frac{a}{b}\right)^2+\left(\frac{a}{c}\right)^2+\left(\frac{b}{a}\right)^2+1+\left(\frac{b}{c}\right)^2+\left(\frac{c}{a}\right)^2+\left(\frac{c}{b}\right)^2+1\)
\(P=3+\left(\frac{a}{b}\right)^2+\left(\frac{b}{a}\right)^2+\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2+\left(\frac{c}{a}\right)^2+\left(\frac{c}{b}\right)^2\)
\(P\ge3+2\sqrt{\frac{\left(ab\right)^2}{\left(ab\right)^2}}+2\sqrt{\frac{\left(ab\right)^2}{c^4}}+2\sqrt{\frac{c^4}{\left(ab\right)^2}}\)
\(P\ge5+2\left(\frac{ab}{c^2}+\frac{c^2}{ab}\right)=5+2\left(\frac{ab}{c^2}+\frac{c^2}{16ab}+\frac{15c^2}{ab}\right)\)
\(P\ge5+2\left(2\sqrt{\frac{abc^2}{16abc^2}}+\frac{15}{16}.4\right)=\frac{27}{2}\)
\(\Rightarrow P_{min}=\frac{27}{2}\) khi \(2a=2b=c\)
