Đơn giản biểu thức ta được:
\(P=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)\)
\(\Leftrightarrow P=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right).\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\)
\(\Leftrightarrow P=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right).\dfrac{\left(x-1\right)\left(y-1\right)}{xy}\)
\(\Leftrightarrow P=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right).\dfrac{\left(-x\right)\left(-y\right)}{xy}\)
\(\Leftrightarrow P=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(\Leftrightarrow P=\left(1+\dfrac{1}{xy}\right)+\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Leftrightarrow P=1+\dfrac{1}{xy}+\dfrac{x+y}{xy}\)
\(\Leftrightarrow P=1+\dfrac{1}{xy}+\dfrac{1}{xy}\)
\(\Leftrightarrow P=1+\dfrac{2}{xy}\)
Ta bắt đầu tìm \(MIN\):
Áp dụng BĐT: \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow P\ge1+\dfrac{2}{\dfrac{1}{4}}=9\)
\(MIN_P=9\) xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
