Đáp án:
$\min P = 9\Leftrightarrow x = y =\dfrac12$
Giải thích các bước giải:
$P =\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)$
$\to P =\dfrac{x^2 -1}{x^2}\cdot\dfrac{y^2-1}{y^2}$
$\to P =\dfrac{(x-1)(x+1)(y-1)(y+1)}{x^2y^2}$
$\to P =\dfrac{(-y)(x+1)(-x)(y+1)}{x^2y^2}$
$\to P =\dfrac{(x+1)(y+1)}{xy}$
$\to P =\dfrac{xy + x + y +1}{xy}$
$\to P =\dfrac{xy +2}{xy}$
$\to P = 1 +\dfrac{2}{xy}$
Ta có:
$(x+y)^2 \geq 4xy$
$\to 1 \geq 4xy$
$\to \dfrac{1}{xy}\geq 4$
$\to \dfrac{2}{xy}\geq 8$
$\to 1 +\dfrac{2}{xy}\geq 9$
$\to P \geq 9$
Dấu $=$ xảy ra $\Leftrightarrow x = y = \dfrac12$
Vậy $\min P = 9\Leftrightarrow x = y =\dfrac12$
