Đáp án:
$\min P = 2018 \Leftrightarrow x =y = 1009$
Giải thích các bước giải:
$\begin{array}{l}\quad P = \dfrac{x^2}{y} + \dfrac{y^2}{x}\\ \to P = \dfrac{x^3 + y^3}{xy}\\ \to P = \dfrac{(x+y)^3 - 3xy(x+y)}{xy}\\ \to P =\dfrac{2018^3}{xy} - 3.2018\\ \text{Ta có:}\\ \quad xy \leq \left(\dfrac{x+y}{2}\right)^2\\ \to xy \leq \dfrac{2018^2}{4}\\ \to \dfrac{1}{xy} \geq \dfrac{4}{2018^2}\\ \to \dfrac{2018^3}{xy} \geq \dfrac{2018^3.4}{2018^2} = 4.2018\\ \to \dfrac{2018^3}{xy} - 3.2018 \geq 4.2018 - 3.2018 = 2018\\ \to P \geq 2018\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x = y = 1009\\ Vậy\,\,\min P = 2018 \Leftrightarrow x =y = 1009\end{array}$
