Question

cho hai số dương x,y có x+y=1

tìm giá trị nhỏ nhất của biểu thức M=\(\left(1-\frac{1}{x^2}\right)\left(1-\frac{1}{y^2}\right)\)

Answer 1

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

=\(\frac{x-1}{x}.\frac{y-1}{y}.\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\) =\(\frac{x-\left(x+y\right)}{x}.\frac{y-\left(x+y\right)}{y}.\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right)\)

=\(\frac{\left(-x\right)\left(-y\right)}{xy}.\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right)\)

=\(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\)

=\(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\ge1+\frac{4}{x+y}+\frac{1}{\frac{\left(x+y\right)^2}{4}}=1+4+\frac{1}{\frac{1}{4}}=9\)

vậy M_min=9 khi x=y=\(\frac{1}{2}\)