Question

Có x>0, y>0. Chứng minh: \(\dfrac{\sqrt{x}}{\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}}\)>_2

Dấu đẳng thức xảy ra khi nào ?

Answer 1

\(\sqrt{\dfrac{x}{y}}-2.\sqrt{\sqrt{\dfrac{x}{y}}}.\sqrt{\sqrt{\dfrac{y}{x}}}+\sqrt{\dfrac{y}{x}}+2.\sqrt{\sqrt{\dfrac{x}{y}}.\sqrt{\dfrac{y}{x}}}\)

=\(\left(\sqrt{\sqrt{\dfrac{x}{y}}}-\sqrt{\sqrt{\dfrac{y}{x}}}\right)^2+2\)

lớn hơn hoặc bằng 2

dấu = xảy ra <=>

\(\left(\sqrt{\sqrt{\dfrac{x}{y}}}-\sqrt{\sqrt{\dfrac{y}{x}}}\right)^2+2=2\)

=>\(\sqrt{\sqrt{\dfrac{x}{y}}}=\sqrt{\sqrt{\dfrac{y}{x}}}\)

=>\(\dfrac{x}{y}=\dfrac{y}{x}\)

=>x²=y²

=>x=y