\(x+y+z=2021xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=2021\)
Ta có:
\(\dfrac{x^2+1+\sqrt{2021x^2+1}}{x}=x+\dfrac{1}{x}+\sqrt{2021+\dfrac{1}{x^2}}=x+\dfrac{1}{x}+\sqrt{\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}+\dfrac{1}{x^2}}\)
\(=x+\dfrac{1}{x}+\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}\le x+\dfrac{1}{x}+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
\(=x+\dfrac{2}{x}+\dfrac{1}{2}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Tương tự:
\(\dfrac{y^2+1+\sqrt{2021y^2+1}}{y}\le y+\dfrac{2}{y}+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\)
\(\dfrac{z^2+1+\sqrt{2021z^2+1}}{z}\le z+\dfrac{2}{z}+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Cộng vế:
\(VT\le x+y+z+\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}=x+y+z+\dfrac{3\left(xy+yz+zx\right)}{xyz}\le x+y+z+\dfrac{\left(x+y+z\right)^2}{xyz}\)
\(VT\le2021xyz+\dfrac{\left(2021xyz\right)^2}{xyz}=2021.2022xyz\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{\dfrac{3}{2021}}\)
