On LHS of the given inequality, we have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = -1,0 as critical points. These critical points divide the real line in three parts as (−∞, −1),[−1,0),[0. ∞).
Case - I :
When −∞ < x < −1
|x + 1| + |x| > 3
⇒ −x− 1 − x > 3
⇒ x < −2.
Case - II
When −1 ≤ x < 0
x + 1| + |x| > 3
⇒ x + 1 − x > 3 ⇒ 1 > 3
Case - III
When 0 ≤ < ∞,
x + 1| + |x| > 3
⇒ x + 1 + x > 3
⇒ x > 1.
Combining the results of cases (I) (II) and (III), we get,
x ∈ (−∞, −2),∪ [1, ∞)
