`x+y+z+t=2`
`⇔4=[(x+y+z)+t]^2\ge 4(x+y+z)t`
`⇔(x+y+z)t\le 1`
`⇔x+y+z\ge [(x+y)+z]^2t\ge 4(x+y)zt`
`⇔[(x+y+z)(x+y)]/(xyzt)\ge [4(x+y)^2zt]/(xyzt)\ge 16`
Dấu `=` xảy ra $⇔\begin{cases}x+y+z+t=2\\x+y+z=t\\(x+y+z)t=1\\x+y=z\end{cases}⇔\begin{cases}x=y=\dfrac{1}{4}\\ z=\dfrac{1}{2}\\ t=1\end{cases}$
Vậy $Min_B=16⇔\begin{cases}x=y=\dfrac{1}{4}\\ z=\dfrac{1}{2}\\ t=1\end{cases}$