Triangle inequality (for real numbers)

Every real number $x$ satisfies $-|x|\le x\le |x|$ by definition of absolute value. Apply this to both $a$ and $b$: $-|a|\le a\le |a|$ and $-|b|\le b\le |b|$.

Add the two chains of inequalities term by term (allowed since adding preserves direction): $-(|a|+|b|)\le a+b\le |a|+|b|$.

The statement $-(M)\le y\le M$ for $M\ge 0$ is exactly equivalent to $|y|\le M$ by definition of absolute value; here $y=a+b$ and $M=|a|+|b|$, so $|a+b|\le |a|+|b|$, which is $|a+b| \le |a|+|b|$.

Equality requires both chained inequalities to be equalities simultaneously, which forces $a$ and $b$ to have the same sign (both nonnegative or both nonpositive), since only then does no cancellation occur between $a$ and $b$.