ta có \(x^4+y^4+z^4\) \(=\dfrac{1}{2}\left(x^4+\dfrac{y^4+z^4}{2}+y^4+\dfrac{x^4+z^4}{2}+z^4+\dfrac{x^4+y^4}{2}\right)\)
\(\ge\dfrac{1}{2}\left(x^4+y^2z^2+y^4+x^2z^2+z^4+x^2y^2\right)\ge\dfrac{1}{2}\left(2x^2yz+2y^2xz+2z^2yx\right)=xyz\left(x+y+z\right)=3xyz\)
dau = xảy ra khi và chỉ khi \(x=y=z=1\)
\(\Rightarrow\) M=2000+16+1=2017
