Question

cho A=\(\frac{1}{\sqrt{xy}}\). với \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\)tìm max A

Answer 1

6 = \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{2}{\sqrt[4]{xy}}\)

<=> \(\frac{1}{\sqrt{xy}}\le9\)

Vậy max là 9 khi x = y = \(\frac{1}{9}\)