Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[l҉o҉n҉g҉ d҉z҉]

x, y > 0

\(=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\cdot\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]\cdot\frac{\sqrt{xy}\left(x+y\right)}{\left(\sqrt{x^3}+\sqrt{y^3}\right)+\left(x\sqrt{y}+y\sqrt{x}\right)}\)

\(=\frac{2\sqrt{xy}+x+y}{xy}\cdot\frac{\sqrt{xy}\left(x+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{xy}}\cdot\frac{\left(x+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

Ta có : \(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{16}}=\frac{\sqrt{x}+\sqrt{y}}{4}=\frac{\sqrt{x}}{4}+\frac{\sqrt{y}}{4}\ge2\sqrt{\frac{\sqrt{xy}}{4\cdot4}}=2\sqrt{\frac{\sqrt{16}}{16}}=1\)( AM-GM )

Dấu "=" xảy ra <=> x = y = 4 . Vậy MinA = 1