# Reverse Triangle Inequality

Reverse triangle inequality states that the length of each side of a triangle is greater than the difference of the lengths of the other two sides.

Mathematically, it can be expressed as shown:

## For Norms

If x and y are real numbers, and (X, || ⋅ ||) is a normed vector space, then ||x – y|| ≥ | ||x|| – ||y|| |, for all x, y Є X.

### Proof

|x| = |(x – y) + y| and |y| = |(y – x) + x|

Taking norms and applying the triangle inequality for vectors, we have

||x|| = ||(x – y) + y|| ≤ ||x – y|| + ||y|| and ||y|| = ||(y – x) + x|| ≤ ||y – x|| + ||x||

By rearranging, we get

||x|| – ||y|| ≤ ||x – y|| and ||y|| – ||x|| ≤ ||y – x||

Combining the above two, we get the reverse triangle inequality | ||x|| – ||y|| | ≤ ||x – y||

## For Real and Complex Fields

It states that if x and y are either real (ℝ) or complex numbers (ℂ), then |x – y| ≥ | |x| – |y| |, where |x| represents the absolute value of a real or complex number.

## For Metric Spaces

If M = (X, d) is a metric space, then d(x, y) ≥ |d(x, z) – d(y, z)|, for all x, y, z Є X.

### Proof

Let us consider a metric space M = (X, d)

By the triangle’s inequality, we have

d(x, y) + d(y, z) ≥ d(x, z), for all x, y, z Є X

On subtracting d(y, z) from both sides, we get

d(x, y) + d(y, z) – d(y, z) ≥ d(x, z) – d(y, z)

⇒ d(x, y) ≥ d(x, z) – d(y, z) …..(i)

Now, considering the following cases, we get

Case 1. Assuming d(x, z) – d(y, z) ≥ 0, we have

d(x, z) – d(y, z) = |d(x, z) – d(y, z)| …..(ii)

Now, by using (ii) in (i), we get

d(x, y) ≥ |d(x, z) – d(y, z)|

Case 2. Assuming d(x, z) – d(y, z) < 0, we have

d(y, x) + d(x, z) ≥ d(y, z) (by the triangle’s inequality)

⇒ d(x, y) ≥ d(y, z) – d(x, z)

Since d(x, z) – d(y, z) < 0, which implies d(y, z) – d(x, z) > 0

Thus, d(x, y) ≥ d(y, z) – d(x, z)

⇒ d(x, y) ≥ |d(x, z) – d(y, z)| …..(iii)

From (ii) and (iii), we get

d(x, y) ≥ |d(x, z) – d(y, z)|, for all x, y, z Є X.

Hence, the reverse triangle inequality is proved.