Question

Cho các số thực a,y dương. Tìm GTNN của:

A=\(\sqrt{\frac{x^2}{y^2}+\frac{y^2}{x^2}+2}+\frac{\sqrt{xy}}{x+y}\)

Answer 1

\(A=\sqrt{\left(\frac{x}{y}+\frac{y}{x}\right)^2}+\frac{\sqrt{xy}}{x+y}=\frac{x}{y}+\frac{y}{x}+\frac{\sqrt{xy}}{x+y}=\frac{x^2+y^2}{xy}+\frac{\sqrt{xy}}{x+y}\ge\frac{\left(x+y\right)^2}{2xy}+\frac{\sqrt{xy}}{x+y}\)

\(A\ge\frac{\left(x+y\right)^2}{16xy}+\frac{\sqrt{xy}}{2\left(x+y\right)}+\frac{\sqrt{xy}}{2\left(x+y\right)}+\frac{7\left(x+y\right)^2}{16xy}\)

\(A\ge3\sqrt[3]{\frac{\left(x+y\right)^2.xy}{16xy.4\left(x+y\right)}}+\frac{7\left(x+y\right)^2}{\frac{16.\left(x+y\right)^2}{4}}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)