ĐK: x≥ 2020; y≥ -2019; z≥2
Có $\sqrt{x-2020}+\sqrt{y+2019}+\sqrt{z-2}=\frac{1}{2}(x+y+z)$
Áp dụng BĐT Cosi
$\sqrt{x-2020}$ ≤ $\frac{x-2020+1}{2}$
$\sqrt{y+2019}$ ≤ $\frac{y+2019+1}{2}$
$\sqrt{z-2}$ ≤ $\frac{z-2+1}{2}$
=> $\sqrt{x-2020}+\sqrt{y+2019}+\sqrt{z-2}≤\frac{x+y+z}{2}$
Mà $\sqrt{x-2020}+\sqrt{y+2019}+\sqrt{z-2}=\frac{1}{2}(x+y+z)$
=> $\frac{1}{2}(x+y+z)≤\frac{1}{2}(x+y+z)$
Dấu "=" xảy ra <=> x- 2020= 1 <=> x=2021 (TM)
y+ 2019= 1 <=> y= -2018 (TM)
z-2= 1 <=> z=3 (TM)
Vậy x= 2021; y=-2018; z=3
