Question

Cho hai số dương x,y thỏa x + y = 1. Tìm GTNN của biểu thức:
\(P=\left(1-\frac{1}{x^2}\right)\left(1-\frac{1}{y^2}\right)\)

Answer 1

\(x+y=1\Rightarrow\left\{{}\begin{matrix}-y=x-1\\-x=y-1\end{matrix}\right.\)

\(P=\frac{\left(x^2-1\right)\left(y^2-1\right)}{x^2y^2}=\frac{\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)}{x^2y^2}=\frac{-y\left(x+1\right)\left(-x\right)\left(y+1\right)}{x^2y^2}\)

\(=\frac{\left(x+1\right)\left(y+1\right)}{xy}=\frac{xy+x+y+1}{xy}=\frac{xy+2}{xy}=1+\frac{2}{xy}\ge1+\frac{8}{\left(x+y\right)^2}=9\)

\(P_{min}=9\) khi \(x=y=\)