x+y=1⇒x=1−y;y=1−x⇒A=x√y+y√x
đặt: \sqrt{x}=a;\sqrt{y}=b\Rightarrow a^2+b^2=1
\Rightarrow A=\frac{a^2}{b}+\frac{b^2}{a}=\frac{a^4}{a^2b}+\frac{b^4}{ab^2}\geq \frac{(a^2+b^2)^2}{a^2b+ab^2} ( cô - si)
\Rightarrow A\geq \frac{1}{a^2b+ab^2}
có:a^2b+ab^2=ab(a+b)\leq \frac{a^2+b^2}{2\sqrt{2(a^2+b^2)}}
\Rightarrow A\geq \frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}
\Rightarrow MinA=........; dấu "=" xảy ra......
Tự lm nốt nhé