Question

cho x,y hoa mãn \(x>y,xy=1\) ,A=\(\dfrac{x^2+y^2}{x-y}\) minA=?

Answer 1

Lời giải:

Ta có:

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

\(=\frac{(x-y)^2+2xy}{x-y}=\frac{(x-y)^2+2}{x-y}\) (do \(xy=1\) )

\(=x-y+\frac{2}{x-y}\)

Áp dụng BĐT Cauchy cho 2 số \(x-y, \frac{2}{x-y}\) dương ta có:

\(A=(x-y)+\frac{2}{x-y}\geq 2\sqrt{(x-y).\frac{2}{x-y}}=2\sqrt{2}\)

Vậy \(A_{\min}=2\sqrt{2}\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-y=\sqrt{2}\\ xy=1\end{matrix}\right.\) \(\Leftrightarrow (x,y)=\left(\frac{\sqrt{6}+\sqrt{2}}{2}; \frac{\sqrt{6}-\sqrt{2}}{2}\right)\)