# Triangle Inequality

The **triangle inequality** states that the sum of the length of any two sides of a triangle is greater than the length of the remaining side.

It follows from the fact that a straight line is the shortest path between two points. The inequality is strict if the triangle is non-degenerate (meaning it has a non-zero area).

#### Contents

## Examples

## If the two sides of a non-degenerate triangle are equal to \(6\), what are the possible values for the length of the third side?

Let the third side have length \(X\), then from the triangle inequality we have \(6+6>X\) and \(6+X>6\), implying \(0<X<12.\) So any real number \(X\) such that \(0<X<12\) is a possible length of the third side. \(_\square\)

## If the lengths of a non-degenerate triangle are 9 and 15, what are the possible lengths of the third side?

Let the third side have length \( x > 0\). The 3 sides must satisfy \[\begin{array} &x < 9 + 15, &15 < x + 9, &9 < x + 15.\end{array}\] The third inequality is trivially true, and the first two give \( 6 < x < 24 \). \(_\square\)

## If the three integer sides of a triangle are given by \(x+2\), \(2x+7\), \(4x+1\), what is the greatest possible value of \(x\)?

By the triangle inequality we have

\[\begin{align} (x+2)+(2x+7)>(4x+1) &\Rightarrow x<8\\ (x+2)+(4x+1)>(2x+7) &\Rightarrow x>\frac{4}{3}\\ (2x+7)+(4x+1)>(x+2) &\Rightarrow x>-\frac{6}{5}, \end{align}\]

which implies \(\displaystyle{\frac{4}{3}<x<8}.\) So the greatest possible integer value of \(x\) is \(7\). \(_\square\)

## Given \(5\) sticks of respective lengths \(1,3,5,9\) and \(10\), how many distinct triangles can be formed from three of the sticks?

If one side of the triangle has a side length of \(1\), then there is no possible combination to form a triangle. In fact, if the three sides of a triangle have distinct integer lengths, then it is impossible to have one side of unit length.

If one side has a length of \(3\), the only possible combination is \((3,9,10)\).

If one side has a length of \(5\), the only possible combination is \((5,9,10)\).

If one side has a length of \(9\), the possible combinations are \((9,3,10)\) and \((9,5,10)\).

If one side has a length of \(10\), the possible combinations are \((10,9,3)\) and \((10,9,5)\).Since changing the order of the sides doesn't change the triangle itself, we can form \(2\) distinct triangles. \(_\square\)

I was cleaning up my attic recently and found a set of at least 14 sticks which a curious Italian gentleman sold me some years ago. Trying hard to figure out why I bought it from him, I realized that the set has the incredible property that there are no \(3\) sticks that can form a triangle. If the set has two sticks of length \(1\), which are the smallest, what is the least possible length of the \({ 14 }^\text{th}\) stick?

## Vectors

The triangle inequality has the following formulation in terms of vectors in \( {\mathbb R}^n\):

Let \( \mathbf x\) and \( \mathbf y\) be vectors in \({\mathbb R}^n.\) Let \( \|\cdot\| \) denote the norm of a vector. Then \[ \|{\mathbf x}+{\mathbf y}\| \le \|{\mathbf x}\| + \|{\mathbf y}\|.\]

Square both sides and subtract: \[ \begin{align} \left( \|{\mathbf x}\| + \|{\mathbf y}\| \right)^2 - \|{\mathbf x}+{\mathbf y}\|^2 &= \|{\mathbf x}\|^2 + \|{\mathbf y}\|^2 + 2\|{\mathbf x}\|\|{\mathbf y}\| - ({\mathbf x}+{\mathbf y}) \cdot ({\mathbf x}+{\mathbf y}) \\ &= \|{\mathbf x}\|^2 + \|{\mathbf y}\|^2 + 2\|{\mathbf x}\|\|{\mathbf y}\| - ({\mathbf x} \cdot {\mathbf x}) - ({\mathbf y} \cdot {\mathbf y}) - 2({\mathbf x}\cdot {\mathbf y}) \\ &= 2\|{\mathbf x}\|\|{\mathbf y}\| - 2({\mathbf x}\cdot {\mathbf y}) \\ &\ge 0 \end{align} \] by the Cauchy-Schwarz inequality.

Note that equality holds in Cauchy-Schwarz if and only if the vectors are parallel, which is the condition that the triangle be degenerate.

The above proof is essentially the same as using the law of cosines: if \( a,b,c\) are the sides of a triangle, \[ c^2 = a^2+b^2-2ab\cos(\theta), \] and the right side is maximized when \( \cos(\theta)\) is minimized, i.e. when it is \( -1,\) so \[ c^2 \le a^2+b^2-2ab(-1) = (a+b)^2, \] so \( c \le a+b,\) with equality if and only if the angle between the sides of lengths \(a\) and \( b\) is \( 180^{\circ}.\)

## Metric spaces

Main article: Metric space

The triangle inequality is a fundamental property of generalized distance functions called **metrics**, which are used to construct metric spaces. A metric is a function \(d(x,y)\) which takes two arguments from a set \(X\) and produces a nonnegative real number, with the following properties:

\(d(x,y) = 0 \) if and only if \( x=y\)

\(d(x,y)=d(y,x)\)

\(d(x,y)+d(y,z) \ge d(x,z)\)

This third property is the triangle inequality, and it is important in proving many of the topological properties of metric spaces.

**Cite as:**Triangle Inequality.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/triangle-inequality/