Lời giải:
Ta có: \(A=\left(1-\frac{4}{x^2}\right)\left(1-\frac{4}{y^2}\right)\)
\(A=\frac{(x^2-4)(y^2-4)}{x^2y^2}\)
\(A=\frac{[x^2-(x+y)^2][y^2-(x+y)^2]}{x^2y^2}=\frac{(-y)(2x+y)(-x)(2y+x)}{x^2y^2}\)
\(A=\frac{xy(2x+y)(2y+x)}{x^2y^2}=\frac{(2x+y)(2y+x)}{xy}=\frac{4xy+2x^2+2y^2+xy}{xy}\)
\(A=5+\frac{2(x^2+y^2)}{xy}=5+\frac{2(x-y)^2+4xy}{xy}=9+\frac{2(x-y)^2}{xy}\)
Thấy rằng \(x,y>0; (x-y)^2\geq 0\Rightarrow \frac{2(x-y)^2}{xy}\geq 0\)
\(\Rightarrow A\geq 9\) hay \(A_{\min}=9\)
Dấu bằng xảy ra khi \(x=y=1\)
\(A=\left(1-\dfrac{4}{x^2}\right)\left(1-\dfrac{4}{y^2}\right)=\left(1-\dfrac{2}{x}\right)\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{2}{y}\right)\left(1+\dfrac{2}{y}\right)\)
\(A=\left(\dfrac{x-2}{x}\right)\left(\dfrac{x+2}{x}\right)\left(\dfrac{y-2}{y}\right)\left(\dfrac{y+2}{x}\right)\)
x+y=2 => x-2 =-y; y-2=-x
\(A=\left(\dfrac{-y}{y}\right)\left(\dfrac{x+2}{x}\right)\left(\dfrac{-x}{x}\right)\left(\dfrac{y+2}{y}\right)=\left(\dfrac{x+2}{x}\right)\left(\dfrac{y+2}{y}\right)\)
\(A=\dfrac{xy+2\left(x+y\right)+4}{xy}=\dfrac{xy+8}{xy}=1+\dfrac{8}{xy}\)
áp dụng co si cho 2 số dương x;y
\(2=\left(x+y\right)\ge2\sqrt{xy}\Rightarrow xy\le1\Rightarrow\dfrac{1}{xy}\ge1\)
\(A\ge1+8.1=9\)đẳng thức x=y=1
