Ta có:
B = ( | x + 3 | + | x - 1 | ) + ( | x + 2 | + | x + 1 | ) + | x + 2 |
\(\ge\left|x+3-x+1\right|+\left|x+2-x-1\right|+\left|x+2\right|\)
= 4 + 1 + | x + 2 | \(\ge\)5
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+3\right)\left(1-x\right)\ge0\\\left(x+2\right)\left(-1-x\right)\ge0\\x+2=0\end{cases}}\Leftrightarrow x=-2\)
Vậy min B = 5 tại x = -2