Đáp án: $P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$
Giải thích các bước giải:
Ta có $x, y\in N^*\to 1\le x,y$
Mà $x+y=2021$
$\to x,y\le 2020$
$\to 1\le x, y\le 2021$
Ta có:
$x+y=2021$
$\to y=2021-x$
$\to xy=x(2021-x)$
$\to P=2021x-x^2$
$\to -P=x^2-2021x$
$\to -P=(x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2$
Mà $1\le x\le 2020$
$\to 1-\dfrac{2021}{2}\le x-\dfrac{2021}{2}\le 2020-\dfrac{2021}{2}$
$\to \dfrac{-2019}{2}\le x-\dfrac{2021}{2}\le \dfrac{2019}{2}$
$\to (x-\dfrac{2021}{2})^2\le \dfrac{2019^2}{4}$
$\to (x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$
$\to -P\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$
$\to P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$
Dấu = xảy ra khi $x=2020$ hoặc $x=1$
