Question

Tìm GTNN của A=\(\left(1+x\right)\left(1+\frac{1}{y}\right)+\left(1+y\right)\left(1+\frac{1}{x}\right)\) với x;y> thỏa mãn \(x^2+y^2=1\)

Answer 1

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Rightarrow\frac{\left(x+y\right)^2}{2}\le1\Rightarrow x+y\le\sqrt{2}\)

\(A=x+\frac{1}{x}+y+\frac{1}{y}+\frac{x}{y}+\frac{y}{x}+2\)

\(A=x+\frac{1}{2x}+y+\frac{1}{2y}+\frac{x}{y}+\frac{y}{x}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)+2\)

\(A\ge2\sqrt{\frac{x}{2x}}+2\sqrt{\frac{y}{2y}}+2\sqrt{\frac{xy}{xy}}+\frac{1}{2}.\frac{4}{\left(x+y\right)}+2\)

\(A\ge4+2\sqrt{2}+\frac{2}{x+y}\ge4+2\sqrt{2}+\frac{2}{\sqrt{2}}=4+3\sqrt{2}\)

\(\Rightarrow A_{min}=4+3\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)