Ta có
$M = \dfrac{4x^2 - 3x + 1}{4x + 2020}$
$= \dfrac{4x^2 + 2020x - 2023x - 1021615 + 1021616}{4x + 2020}$
$= x - \dfrac{2023}{4} + \dfrac{255404}{x + 505}$
$= x + 505 + \dfrac{255404}{x+505} -\dfrac{4043}{4}$
Áp dụng BĐT Cauchy ta có
$( x + 505) + \dfrac{255404}{x+505} \geq 2 \sqrt{(x+505) . \dfrac{255404}{x+505}} = 4\sqrt{63851}$
$<-> x + 505 + \dfrac{255404}{x+505} -\dfrac{4043}{4} \geq 4\sqrt{63851} - \dfrac{4043}{4}$
Dấu "=" xảy ra khi $(x+505)^2 = 255404$ hay $x = 2 \sqrt{6318} - 505$
Vậy GTNN của M là $4\sqrt{63851} - \dfrac{4043}{4}$ khi $x = 2 \sqrt{6318} - 505$.
