## Proof of the Triangle Inequality for Real Numbers

The triangle inequality for real numbers is

$ |a+b| \le |a| + |b|$

in which $a$ is a variable for a real number, and $b$ is a variable for a real number.

*Proof:*

I use four cases.

Case 1: If $a=0$ and $b$ is any real number, then the left side of the triangle inequality is $|b|$. And the right side is $|b|$. So the left side is equal to the right side. Similarly, if $b=0$ and $a$ is any real number, then the left side is $|a|$ and the right side is $|a|$.

Case 2: If $a$ and $b$ have opposite signs, $a+b=b+a$ is a *subtraction* of a positive number from another positive number, which has an absolute value *less* than the sum of the absolute value of the first number with the absolute value of the second number. This is the origin of the $<$ in the triangle inequality.

Case 3: If $a$ and $b$ are positive numbers, then $a+b > 0$, which means that $|a+b|=a+b$. This is the left side of the triangle inequality. Since $a$ is positive, $a=|a|$, and similarly $b=|b|$ since $b$ is positive. Hence $|a|+|b|=a+b$ is the right side. Therefore, $ |a+b| = a+b = |a| + |b|$.

Case 4: If $a$ and $b$ are negative numbers, then $a+b$ is the addition of two negative numbers. Suppose for the sake of argument that $a = -A$ and $b = -B$ in which $A$ and $B$ are *positive* real numbers. Then $|a+b|=|-A-B|=|(-1)(A+B)|=|A+B|$ from this absolute value property. So the left side is actually $|A+B| = A+B$ since $A+B$ is a positive real number. Next, the right side of the triangle inequality is $|a| + |b| = |-A| + |-B| = A + B$ by the definition of the absolute value. Thus the left side is equal to the right side since $ |a+b| = A+B = |a| + |b|$.

$\square$