Question

Cho \(x+y\le1\). Tìm GTNN của \(A=x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}+4\)

Answer 1

\(A\ge\frac{1}{2}\left(x+y\right)^2+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)^2+4\)

\(A\ge\frac{1}{2}\left(x+y\right)^2+\frac{8}{\left(x+y\right)^2}+4=\frac{1}{2}\left(x+y\right)^2+\frac{1}{2\left(x+y\right)^2}+\frac{15}{2\left(x+y\right)^2}+4\)

\(A\ge2\sqrt{\frac{\left(x+y\right)^2}{4\left(x+y\right)^2}}+\frac{15}{2.1}+4=\frac{25}{2}\)

\(A_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)