Prove that $\neg \exists x$ such that $ \lvert x-1\rvert+\lvert x+1 \rvert <1$

Suppose there exists $x_0\in \Bbb R $ such that $\lvert x_0-1\rvert+\lvert x_0+1 \rvert <1$. Then \begin{align}|2|&=|x_0-1-(x_0+1)|\\ &\leq \lvert x_0-1\rvert+\lvert -(x_0+1) \rvert\\&=\lvert x_0-1\rvert+\lvert x_0+1 \rvert\\ & <1, \end{align} a contradiction .

Case 1: $x \le -1$

$|x - 1| + |x + 1| = -(x - 1) + -(x + 1) = -2x \ge 2$

Case 2 : $-1 \le x \le 1$

$|x - 1| + |x + 1| = -(x - 1) + (x + 1) = 2$

Case 3 : $1 \le x$

$|x - 1| + |x + 1| = (x - 1) + (x + 1) = 2x \ge 2$

Because by the triangle inequality $$|x+1|+|x-1|=|x+1|+|1-x|\geq|x+1+1-x|=2\geq1.$$