# Prove that $\neg \exists x$ such that $\lvert x-1\rvert+\lvert x+1 \rvert <1$ [duplicate]

In Spivak Calculus, chapter 1 question 11 vi. asks the reader to find all numbers $$x$$ for which $$\lvert x-1\rvert+\lvert x+1 \rvert <1$$. Intuitively speaking, it is quite obvious that there is no number $$x$$ that would make this inequality true, and in checking a graph of the equation I am now certain of this. Despite the textbook not asking for a proof, I am curious as to how one would go about proving this rigorously? Would this require using multiple cases or is it possible without?

Thanks for any insight.

Note: Ideally any proof presented would not require calculus, as that is beyond the scope of the first chapter of the book, but I'd still be interested in any proof involving calculus as this is purely out of curiosity.

## marked as duplicate by Martin R, Gibbs, Vinyl_cape_jawa, Song, N. F. TaussigMar 13 at 13:27

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 $$|x+1|+|x-1|=|x+1|+|1-x|\geq|x+1+1-x|=2\geq1.$$