\(P=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2016}{ab+bc+ca}\)
\(P\ge\frac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\frac{2016}{\frac{1}{3}\left(a+b+c\right)^2}\)
\(P\ge\frac{6057}{\left(a+b+c\right)^2}\ge\frac{6057}{3^2}=673\)
Dấu "=" xảy ra khi \(a=b=c=1\)