Question

Cho ba số thực x,y,thay đổi thỏa mãn \(\left(x+y\right)xy=x^2+y^2-xy\)

tìm Max    \(A=\frac{1}{x^3}+\frac{1}{y^3}\)

Answer 1

Ta có \(\left(x+y\right)xy=x^2-xy+y^2\)

=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)

MÀ \(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)^2,\frac{1}{xy}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^{^2}\)

=> \(\frac{1}{x}+\frac{1}{y}\le4\)

\(A=\frac{1}{x^3}+\frac{1}{y^3}=\frac{x^3+y^3}{x^3y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)

Vậy MaxA=16 khi x=y=1/2