Question

Cho x,y > 0 thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

Tìm GTNN : A= \(\sqrt{x}+\sqrt{y}\)

Answer 1

Áp dụng BĐT Cauchy , ta có :

\(\sqrt{x}+\sqrt{x}+\dfrac{1}{x}\ge3\sqrt[3]{\sqrt{x}.\sqrt{x}.\dfrac{1}{x}}=3\)

\(\sqrt{y}+\sqrt{y}+\dfrac{1}{y}\ge3\sqrt[3]{\sqrt{y}.\sqrt{y}.\dfrac{1}{y}}=3\)

\(\Rightarrow2\left(\sqrt{x}+\sqrt{y}\right)+\dfrac{1}{x}+\dfrac{1}{y}\ge6\)

\(\Leftrightarrow2\left(\sqrt{x}+\sqrt{y}\right)\ge4\)

\(\Leftrightarrow\sqrt{x}+\sqrt{y}\ge2\)

\(\Rightarrow A_{Min}=2."="\Leftrightarrow x=y=1\)