A = \(2x^2+\frac{6}{x^2}+3y^2+\frac{8}{y^2}=2x^2+\frac{2}{x^2}+\frac{4}{x^2}+3y^2+\frac{3}{y^2}+\frac{5}{y^2}=\)
Ta có : \(\frac{4}{x^2}+\frac{5}{y^2}\ge9\)
\(2x^2+\frac{2}{x^2}\ge2\sqrt{\frac{2x^2.2}{x^2}=4}\)
\(3y^2+\frac{3}{y^2}\ge6\)
=> \(A\ge19\)