Đáp án: $P\ge 2039$
Giải thích các bước giải:
Ta có:
$P=3x+\dfrac{6}{x}+\dfrac{8}{y}+2y+2020$
$\to P=3x+\dfrac{6}{x})+(\dfrac{8}{y}+2y+2020$
$\to P=(\dfrac{3x}{2}+\dfrac{6}{x})+(\dfrac{y}{2}+\dfrac{8}{y})+(\dfrac32x+\dfrac32y)+2020$
$\to P\ge 2\sqrt{\dfrac{3x}{2}\cdot\dfrac{6}{x}}+2\sqrt{\dfrac{y}{2}\cdot\dfrac{8}{y}}+\dfrac32(x+y)+2020$
$\to P\ge 2\sqrt{\dfrac{3x}{2}\cdot\dfrac{6}{x}}+2\sqrt{\dfrac{y}{2}\cdot\dfrac{8}{y}}+\dfrac32\cdot 6+2020$
$\to P\ge 2039$
Dấu = xảy ra khi $\dfrac32x=\dfrac6x, \dfrac{y}2=\dfrac8y, x+y=6$
$\to x=2,y=4$
