Prove that $|ab+1|>|a+b|$ with $|a|<1$, $|b|<1$

Question

Prove that $|ab+1|>|a+b|$ with $|a|<1$, $|b|<1$

$a$, $b$ are real numbers

Where $|a|$ is the absolute value of $a$.

Every time, I arrive to a dead-end.

Answer 1

Hint: Since $a<1$ and $b<1$, we have $$(a-1)(b-1)>0.$$ Similarly $$(a+1)(b+1)>0.$$ Expanding both of them gives you the answer.

Answer 2

Hint: $$(ab+1)^2-(a+b)^2=(ab-a-b+1)(ab+a+b+1)=(a-1)(b-1)(a+1)(b+1)=(a^2-1)(b^2-1)>0$$