Đáp án: $9$
Giải thích các bước giải:
Ta có:
$P=(1-\dfrac{1}{x^2})(1-\dfrac{1}{y^2})$
$\to P=\dfrac{x^2-1}{x^2}\cdot\dfrac{y^2-1}{y^2}$
$\to P=\dfrac{(x-1)(x+1)}{x^2}\cdot \dfrac{(y-1)(y+1)}{y^2}$
$\to P=\dfrac{(1-x)(x+1)}{x^2}\cdot \dfrac{(1-y)(y+1)}{y^2}$
Mà $x+y=1\to 1-x=y, 1-y=x$
$\to P=\dfrac{y(x+1)}{x^2}\cdot \dfrac{x(y+1)}{y^2}$
$\to P=\dfrac{xy(x+1)(y+1)}{x^2y^2}$
$\to P=\dfrac{(x+1)(y+1)}{xy}$
$\to P=\dfrac{xy+x+y+1}{xy}$
$\to P=\dfrac{xy+1+1}{xy}$
$\to P=\dfrac{xy+2}{xy}$
$\to P=1+\dfrac{2}{xy}$
$\to P=1+\dfrac{8}{4xy}$
$\to P\ge 1+\dfrac{8}{(x+y)^2}$
$\to P\ge 1+\dfrac{8}{1}$
$\to P\ge 9$
Dấu = xảy ra khi $x=y=\dfrac12$
