\(A=\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\)
\(>=\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+ac+bc}\)(bđt svacxo)\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+ac+bc}+\frac{1}{ab+ac+bc}+\frac{7}{ab+ac+bc}\)
\(>=\frac{9}{a^2+b^2+c^2+ab+ac+bc+ac+ac+bc}+\frac{7}{ab+ac+bc}\)(bđt svacxo)
\(=\frac{9}{a^2+b^2+c^2+2ab+2ac+2bc}+\frac{7}{ab+ac+bc}=\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+ac+bc}\)
\(=\frac{9}{1}+\frac{7}{ab+ac+bc}=9+\frac{7}{ab+ac+bc}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc>=ab+ac+bc+2ab+2ac+2bc\)
\(=3ab+3ac+3bc=3\left(ab+ac+bc\right)\Rightarrow\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\cdot1=\frac{1}{3}>=ab+ac+bc\Rightarrow ab+ac+bc< =\frac{1}{3}\)
\(\Rightarrow9+\frac{7}{ab+ac+bc}>=9+\frac{7}{\frac{1}{3}}=9+7\cdot3=9+21=30\)
\(\Rightarrow A>=30\)dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
vậy min A là 30 khi \(a=b=c=\frac{1}{3}\)
