Áp dụng BDT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|x+3\right|=\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|=4\)
Dấu "=" xảy ra khi (1-x)(x+3)\(\ge0\)\(\Leftrightarrow-3\le x\le1\)
Vậy \(-3\le x\le1\)