Ta có :\(\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ca}\)
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2007}{ab+bc+ca}\)
Áp dụng bđt Cauchy - Schwarz dạng Engel ta có :
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{3^2}=1\)( do \(a+b+c\le3\)) (1)
Lại có : \(a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
nên \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)\ge3\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\Leftrightarrow9\ge3\left(ab+bc+ac\right)\Rightarrow ab+bc+ac\le3\)
\(\Rightarrow\frac{2007}{ab+bc+ac}\ge\frac{2007}{3}=669\)(2)
Từ (1) ; (2) \(\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2007}{ab+bc+ca}\ge670\)
Hay \(\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ca}\ge670\)(đpcm)
