Question

cho \(x+y\le1\). Tìm giá trị nhỏ nhất của biểu thức P

\(\left(1+\dfrac{1}{x^2}\right)\left(1+\dfrac{1}{y^2}\right)\)

Answer 1

Ta có: \(4xy\le\left(x+y\right)^2\le1\)

\(\Leftrightarrow xy\le\dfrac{1}{4}\)

\(A=\left(1+\dfrac{1}{x^2}\right)\left(1+\dfrac{1}{y^2}\right)\)

\(=\left(1+\dfrac{1}{4x^2}+\dfrac{1}{4x^2}+\dfrac{1}{4x^2}+\dfrac{1}{4x^2}\right)\left(1+\dfrac{1}{4y^2}+\dfrac{1}{4y^2}+\dfrac{1}{4y^2}+\dfrac{1}{4y^2}\right)\)

\(\ge5\sqrt[5]{\dfrac{1}{4^4x^8}}.5\sqrt[5]{\dfrac{1}{4^4y^8}}\)

\(=25\sqrt[5]{\dfrac{1}{4^8}.\dfrac{1}{\left(xy\right)^8}}\ge25\sqrt[5]{\dfrac{1}{4^8}.\dfrac{1}{\left(\dfrac{1}{4}\right)^8}}=25\)