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 […]