Đặt a= 1-x
b=1-y
c=1-z
\(\Rightarrow\) a+b+c= 1-x+1-y+1-z=0 và ;b;c=[-1;1]
khi đó A=(1-a)^4 + (1-b)^4 + (1-c)^4 + 12abc
=3-4(a+b+c) + 6 ( \(a^2+b^2+c^2\))-\(4\left(a^3+b^3+c^3\right)+a^4+b^4+c^4+12abc\)
=\(3+6\left(a^2+b^2+c^2\right)-4.3abc-12abc\) do\(\left(a^3+b^3+c^3=abc\right)\)
=\(3+6\left(a^2+b^2+c^2\right)+a^4+b^4+c^4\ge3\)
dấu bằng xảy ra khi a=b=c=0
\(\Leftrightarrow\)x=y=z=1