Question

cho x,y là hai số dương thỏa mãn x + y = 1, tìm min

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

Answer 1

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

\(P=\frac{xy+x+y+1}{xy}=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\)

\(P\ge1+\frac{4}{x+y}+\frac{4}{\left(x+y\right)^2}=9\)

\(P_{min}=9\) khi \(x=y=\frac{1}{2}\)