\(1=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{3}{\sqrt[3]{a^2b^2c^2}}\Rightarrow\sqrt[3]{a^2b^2c^2}\ge3\Rightarrow a^2b^2c^2\ge27\)
\(A=1+a^2b^2c^2+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2\)
\(A\ge1+27+3\sqrt[3]{a^2b^2c^2}+3\left(\sqrt[3]{a^2b^2c^2}\right)^2\)
\(A\ge1+27+3.3+3.3^2=...\)
Dấu "=" xảy ra khi \(a=b=c=...\)