Điều kiện: \(x;y>1\)
\(A=\dfrac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\)
\(\ge\dfrac{\left(x+y\right)^2}{x+y-2}\)
Đặt \(x+y=a\left(a>2\right)\)
\(\Rightarrow A=\dfrac{a^2}{a-2}=\dfrac{8\left(a-2\right)+\left(a^2-8a+16\right)}{a-2}=8+\dfrac{\left(a-4\right)^2}{a-2}\ge8\)
Dấu "=" xảy ra khi x = y = 2
Vậy \(Min_A=8\Leftrightarrow x=y=2\)
Đúng 0
Bình luận (3)
