\(9=3a^2+2b^2+2c^2+2bc\)
\(\Leftrightarrow9=\left(a+b+c\right)^2+2a^2+b^2+c^2-2a\left(b+c\right)\)
\(\Leftrightarrow9\ge\left(a+b+c\right)^2+2a^2+\frac{1}{2}\left(b+c\right)^2-2a\left(b+c\right)\)
\(\Leftrightarrow9\ge\left(a+b+c\right)^2+\frac{1}{2}\left(2a-b-c\right)^2\ge\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le3\)
Ta có:
\(P=a+b+c+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge a+b+c+\frac{18}{a+b+c}\)
\(P\ge a+b+c+\frac{9}{a+b+c}+\frac{9}{a+b+c}\)
\(P\ge2\sqrt{\frac{9\left(a+b+c\right)}{a+b+c}}+\frac{9}{3}=9\)
Dấu "=" xảy ra khi \(a=b=c=1\)