\(A=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)=\left(1-\dfrac{1}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\left(1-1-\dfrac{y}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-1-\dfrac{x}{y}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\dfrac{1}{y}+\dfrac{1}{x}+1+\dfrac{1}{xy}=\dfrac{x+y+1}{xy}+1=\dfrac{2}{xy}+1\)
áp dụng bđt cosi \(x+y\ge2\sqrt{xy}< =>1\ge2\sqrt{xy}< =>xy\le\dfrac{1}{4}< =>\dfrac{2}{xy}\ge8\)
<=> A >= 8+1=9
dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
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