Question

Prove the triangle inequality, which states that if \(x\) and \(y\) are real numbers, then \(\left| x \right| + \left| y \right| \ge \left| {x + y} \right|\) (where \(\left| x \right|\) represents the absolute value of \(x\),which equals \(x\) if \(x \ge 0\) and equals \( - x\) if \(x < 0\)).

Short answer

The triangle inequality;

\(\left| x \right| + \left| y \right| \ge \left| {x + y} \right|\) follows, whenever \(x\) and \(y\) are real numbers.

Step-by-step solution

Introduction

It is known that \(\left| x \right|\)represents the absolute value of \(x\) if \(x \ge 0\).

The purpose is to prove the triangle inequality.

The triangle inequality is;

\(\left| x \right| + \left| y \right| \ge \left| {x + y} \right|\), whenever \(x\) and \(y\) are real numbers.

Case 1: \(x \ge 0\) and \(y \ge 0\)

Here,

\(\begin{array}{c}\left| x \right| + \left| y \right| = x + y\\ = \left| {x + y} \right|\end{array}\)

Case 2: \(x < 0\)and \(y < 0\)

Here,

\(\begin{array}{c}\left| x \right| + \left| y \right| = - x - y\\ = - \left( {x + y} \right)\\ = \left| {x + y} \right|\end{array}\)

This is because\(\left( {x + y} \right) < 0\).

Case 3: \(x \ge 0\) and \(y < 0\)

Here,

\(\begin{array}{c}\left| x \right| + \left| y \right| = x - y\\ = x + \left( { - y} \right)\end{array}\)

Sub case 1: \(x \ge - y\)

This implies,

\(\left| x \right| + \left| y \right| = x + y\)

But \(y < 0\)

This implies,

\( - y > y\)

This implies,

\(\begin{array}{c}\left| x \right| + \left| y \right| = x + \left( { - y} \right)\\ > x + y\\ = \left| {x + y} \right|\end{array}\)

Sub case 2: \(x < - y\)

This implies,

\(\begin{array}{c}\left| {x + y} \right| = - \left( {x + y} \right)\\ = - x + \left( { - y} \right)\end{array}\)

As \(x \ge 0\)

So,

\(x > - x\).

This implies,

\(\begin{array}{c}\left| x \right| + \left| y \right| = x + \left( { - y} \right)\\ \ge - x + \left( { - y} \right)\\ = \left| {x + y} \right|\end{array}\)

Case 4: \(x < 0\) and \(y \ge 0\)

Interchange the roles of \(x\) and \(y\) in case 3,

\(\begin{array}{c}\left| x \right| + \left| y \right| = \left( { - x} \right) + y\\ \ge x + y\\ = \left| {x + y} \right|\end{array}\)

Thus,

\(\left| x \right| + \left| y \right| \ge \left| {x + y} \right|\), whenever \(x\)and \(y\)are real numbers.

Hence, proved.