Cho số thực x, y > 1. Min $\dfrac{(x^3 + y^3) - (x^2 + y^2)}{(x - 1)(y - 1)}$?

V

vipboycodon

$A = \dfrac{x^3+y^3-(x^2+y^2)}{(x-1)(y-1)} = \dfrac{x^2}{y-1}+\dfrac{y^2}{x-1} \ge \dfrac{(x+y)^2}{x+y-2}$

Đặt $t = x+y \ (t > 2)$

$\rightarrow A = \dfrac{t^2}{t-2} = t-2+\dfrac{4}{t-2}+4 \ge 8$ (cauchy)

Dấu "=" xảy ra khi $\begin{cases} \dfrac{x}{y-1} = \dfrac{y}{x-1} \\ t = x+y = 4 \end{cases} \leftrightarrow x = y = 2$
 
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