Subtracting 3 from both sides, we get–

|x+1|+|x|–3>0

For this, we have 3 cases,

Case 1:–∞ <x<–1

For this, |x+1|=–(x+1) and |x|=–x

–(x+1)–x–3>0

–2x–1–3>0

2x+4<0

x<–2

⇒ x ϵ (–∞ , –2) …(1)

Case 2: –1<x<0

For this, |x+1|=x+1 and |x|=–x

x+1–x–3>0

–2>0

Which is absurd, thus it leads to no solution.

Case 3: 0<x<∞

For this, |x+1|=x+1 and |x|=x

x+1+x–3>0

2x–2>0

x>1

⇒ x ϵ (1 , ∞ ) …(2)

⇒ xϵ (–∞ , –2)⋃ (1 , ∞ ) (from 1 and 2)

We can verify the answers using graph as well.