Theo BĐT Cosi ta có: \(\hept{\begin{cases}\frac{x^4+y^4}{2}\ge\sqrt{x^4\cdot y^4}=x^2y^2\\\frac{y^4+z^4}{2}\ge\sqrt{y^4\cdot z^4}=y^2z^2\\\frac{z^4+x^4}{2}\ge\sqrt{z^4\cdot x^4}=x^2z^2\end{cases}\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2}\)
chứng minh tương tự: \(x^2y^2+y^2z^2+z^2x^2\ge xy^2z+xyz^2+x^2yz\Leftrightarrow x^2y^2+y^2z^2+x^2z^2\ge xyz\left(x+y+z\right)\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2\ge3xyz\)(do x+y+z=3)
Do đó: \(x^4+y^4+z^4\ge3xyz\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x^4=y^4;y^4=z^4;z^4=x^4\\x^2y^2=y^2z^2;y^2z^2=z^2x^2;z^2x^2=x^2y^2\end{cases}\Leftrightarrow x=y=z}\)(1)
mà x+y+z=3 (2)
Từ (1) và (2) => 3x=3 => x=1 => y=z=1
=> \(x^{2018}+y^{2019}+x^{2020}=1+1+1=3\)
