Question

Cho x, y lớn hơn 0 thỏa mãn \(x+y\le1\) . tìm min của \(M=xy+\frac{9}{xy}\)

Answer 1

chào bn

Answer 2

 bđt cauchy \(xy+\frac{9}{xy}\ge2\sqrt{xy\cdot\frac{9}{xy}}=18\)

\(xy+\frac{9}{xy}\ge2\sqrt{xy\cdot\frac{9}{xy}}=18\)

 

Answer 3

Còn điều kiện x + y <= 1 coi chớ sai rồi kìa Tiểu Nghé

Answer 4

xy + \(\frac{9}{xy}\)\(\ge\)xy + \(\frac{9\left(x+y\right)}{xy}\)= xy + \(\frac{9}{x}+\frac{9}{y}\)\(\ge\)\(9\sqrt[3]{3}\)

Answer 5

Có: \(\frac{1}{4}\ge\left(\frac{x+y}{2}\right)^2\ge\left(\frac{2\sqrt{xy}}{2}\right)^2=xy\)

\(M=\left(xy+\frac{1}{16xy}\right)+\frac{143}{16xy}\ge2\sqrt{xy.\frac{1}{16xy}}+\frac{143}{16.\frac{1}{4}}=\frac{145}{4}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)