Question

Cho hai số dương x,y và x +y = 1 .Tìm GTNn của M = \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\)

Answer 1

Áp dụng BĐT AM-GM ta có:

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

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

\(=x^2y^2+\dfrac{1}{x^2y^2}+2=x^2y^2+\dfrac{1}{256x^2y^2}+\dfrac{255}{256x^2y^2}+2\)

\(\ge2\sqrt{x^2y^2\cdot\dfrac{1}{256x^2y^2}}+\dfrac{255}{256\cdot\left(xy\right)^2}+2\)

\(\ge2\cdot\dfrac{1}{16}+\dfrac{255}{256\cdot\left(\dfrac{\left(x+y\right)^2}{4}\right)^2}+2\)

\(=\dfrac{1}{8}+\dfrac{255}{256\cdot\left(\dfrac{1}{4}\right)^2}+2=\dfrac{289}{16}\)

Khi \(x=y=\dfrac{1}{2}\)