<s>[INST] Finding all $x \in \mathbb{R}$ with $|x+1| + |x-1| < 4$ We want to find all $x \in \mathbb{R}$ with
$$|x+1| + |x-1| < 4$$
I tried the following, but I'm uncertain if this is correct:
Critical points: $x= -3$ and $x = 1$.
Case $1$:
$$x \in (-\infty, -3) \\ 
|x+1| = -(x+1) = -x-1 \\
|x-1| = -(x-1) = -x+1 \\
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow  \text{ holds } \forall x \in (-\infty,-2[$$
Case $2$:
$$x \in [-3, 1) \\ 
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow  \text{ holds } \forall x \in [-3,-2[$$
Case $3$:
$$x \in (1, \infty) \\ 
|x+1| = x+1 \\
|x-1| = x-1 \\
x+1+x-1 < 4\\
\iff 2x < 4 \\
\iff x < 2 \\ \Rightarrow  \text{ holds } \forall x \in [1,2)$$
The solution would then be $x \in (-\infty,-2) \cup [1,2)$.
Is this correct/wrong? Is this the common approach for questions like these?
 [/INST] Consider a cord of length $4$ with ends attached to the points $-1$ and $1.$ When the cord is maximaly extended to the right or left it reaches the points $2$ and $-2,$ respectively. Therefore the answer is $(-2,2).$
Similarly let the cord of length $2L$ be attached to the points $a<b$ such that $2L>b-a.$ Then the solution of
$$|x-a|+|x-b|<2L$$
is $(-L+{a+b\over 2}, L+{a+b\over 2}).$
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