Our claim is that $|a| + |b| \geq |a + b|$ for real $a$ and $b$ and we want to prove it by contradiction. Hence, we assume that $|a| + |b| < |a + b|$.
Here is a proof.
Case 1: $0 < a < b$.
Because both $a$ and $b$ are greater than zero, this means that $a + b$ is also greater than zero. Then $|a| = a$, $|b| = b$, and $|a + b| = a + b$. We then have $a + b > a + b$ which is a contradiction.
Case 2: $a < 0 < b$ and $|a| < |b|$.
This means that $|a| + |b| > 0$. Because of the conditions, we have $b > a + b \implies |b| > |a + b|$. Since we have $|a| + |b| > 0$, then $|a| + |b| > |b| > |a + b| \implies |a| + |b| > |a + b|$. Contradiction.
Case 3: $a < 0 < b$ and $|a| > |b|$.
This also means that $|a| + |b| > 0$. Because of the conditions, $a < a + b \implies |a| > |a + b|$. Since we have $|a| + |b| > 0$, then $|a| + |b| > |a| > |a + b| \implies |a| + |b| > |a + b|$. Contradiction.
Case 4: $a < b < 0$.
This case can be treated in a similar manner as the first case. Since both $a$ and $b$ are less than zero, this means that $a + b$ is also less than zero. Then $|a| = -a$, $|b| = -b$, and $|a + b| = -a - b$ and we have $-a - b > -a - b$. Contradiction.
Case 5: $a = b$.
This means that $|2a| < |2a|$ which is a contradiction.
Case 6: $a = 0$.
This means that $|b| < |b|$ which is a contradiction.
Because all cases of $a$ and $b$ lead to a contradiction, our assumption must not be true. Therefore, $|a + b| \geq |a + b|$.