Đáp án: $16$
Giải thích các bước giải:
Ta có:
$xy(x+y)=x^2-xy+y^2$
$\to \dfrac{xy(x+y)}{x^2y^2}=\dfrac{x^2-xy+y^2}{x^2y^2}$
$\to \dfrac1x+\dfrac1y=\dfrac1{x^2}-\dfrac1{xy}+\dfrac1{y^2}$
$\to \dfrac1x+\dfrac1y=(\dfrac1x+\dfrac1y)^2-3\cdot \dfrac1{x}\cdot \dfrac1{y}$
$\to \dfrac1x+\dfrac1y\ge (\dfrac1x+\dfrac1y)^2-\dfrac34\cdot (\dfrac1x+\dfrac1y)^2$
$\to \dfrac1x+\dfrac1y\ge \dfrac14\cdot (\dfrac1x+\dfrac1y)^2$
$\to \dfrac1x+\dfrac1y\le 4$
Mặt khác
$\dfrac1x+\dfrac1y=\dfrac1{x^2}-\dfrac1{xy}+\dfrac1{y^2}$
$\to (\dfrac1x+\dfrac1y)^2=(\dfrac1x+\dfrac1y)(\dfrac1{x^2}-\dfrac1{xy}+\dfrac1{y^2})$
$\to (\dfrac1x+\dfrac1y)^2=\dfrac1{x^3}+\dfrac1{y^3}$
$\to A=(\dfrac1x+\dfrac1y)^2\le 16$
$\to GTLN_A=16$
Khi đó $x=y=\dfrac12$
