The Triangle Inequality and Degenerate Triangles

The triangle inequality states that the sum of the lengths of two sides of a triangle is greater than the length of the third side. Equality holds if the triangle is degenerate; that is, the three vertices are collinear.

The triangle inequality is sometimes expressed in terms of vectors as follows: $$|\vec{a}|+|\vec{b}|\ge|\vec{a}+\vec{b}|$$ It is also sometimes expressed in terms of the absolute values of real or complex numbers. This version of the triangle inequality is the same as the vector version, both in appearance and function: $$|a|+|b|\ge|a+b|$$

This is equivalent to the triangle inequality, since one can obviously create a triangle with side lengths $|\vec{a}|$, $|\vec{b}|$, and $|\vec{a}+\vec{b}|$. The triangle is degenerate if $\vec{a}\parallel\vec{b}$, and indeed, if $\vec{a}$ and $\vec{b}$ point in the same direction, equality holds.

So how do we prove the triangle inequality?

Suppose that $BC$ is the longest side of $\triangle ABC$, which may or may not be degenerate.

Drop a perpendicular from $A$ to $BC$. Call the foot of the perpendicular $X$.

By the Pythagorean Theorem, $$AB^2=AX^2+BX^2$$ $$AC^2=AX^2+CX^2$$ Since perfect squares are nonnegative, $$AB^2\ge BX^2$$ $$AC^2\ge CX^2$$ We know that all distances are nonnegative. Therefore, $$AB\ge BX$$ $$AC\ge CX$$ One could also have just used the result that the length of the altitude from a point to a line is the shortest distance to the line. This would have yielded the same result, but it's better to just prove it again.

Since $BC$ is the longest side, we know that $BX+CX=BC$. Thus, we get our result: $$AB+AC\ge BC$$ Since the triangle inequality holds for the longest side of the triangle, it obviously holds for the shorter sides.

One can see that if the triangle is degenerate, then $AX=0$. Then, equality holds since $AB=BX$ and $AC=CX$.

Thus, we have proven the triangle inequality.

The Triangle Inequality states that the sum of any two sides is always greater than the third side. In other words, in any triangle $ABC$, we have the following inequalities: $$AB + AC > BC$$ $$AC + BC > AB$$ $$AB + BC > AC$$

Without loss of generality, let $BC$ be the largest side. Then two of these inequalities are immediately proven.

We already have $BC > AB$ and $BC > AC$, so adding another positive number to the left side still satisfies the inequality. It remains to prove the first inequality, that $AB + AC > BC$.

It is worth noting that the first WLOG assumption is not necessary for the proof. It was included mainly for intuitive purposes, in that if the longest side is already on the greater than side, then the inequality should be obvious. The proof below will only prove the first inequality. However, it can be repeated for all three inequalities without any assumptions on side lengths.

Given any triangle $ABC$, we will construct point $D$ on line $CA$ such that $AD = AB$

In the diagram above, the original triangle is solid while the constructed point and segments are dashed. Since $AD = AB$, we also have $\angle ADB = \angle ABD$. Since $\angle DBC = \angle ABD + \angle ABC$, it is evident that $\angle DBC > \angle ABD = \angle ADB$, or more simply, $\angle DBC > \angle ADB$. Then in triangle $DBC$, the sides opposite the angles in this inequality also satisfy the inequality. In other words, we have $DC > BC$. Recall that we constructed $D$ on line $AC$, so we have $DC = AC + AD = AC + AB$. Hence we have $AB + AC > BC$, as desired.

The proof is complete, but now we have some more intuitive things to go through. Again, take $BC$ to be the longest side. Consider what happens if the inequality doesn't hold: $AB + AC < BC$. Then such a point $A$ doesn't even exist! To see why, let's get rid of $A$ and replace it with $D$ and $E$. In other words, we are given lengths $BD$, $CE$, and $BC$ where $BD+ CE < BC$. Start by constructing $BC$. Then no matter how you draw $BD$ and $CE$, there is no way for them to intersect.

But what if $BD + CE = BC$? Then the only way for $BD$ and $CE$ to intersect if the points $D$ and $E$ coincide on $BC$. In other words, $D$ and $E$ are essentially the same point, so we'll reintroduce $A$ to be that point. Furthermore, $B$, $A$, and $C$ are collinear (lie on the same line) with $A$ in between $B$ and $C$. This should easily make sense since we're basically dividing the segment $BC$ into two segments: $AB$ and $AC$. Then the lengths should add up: $AB + AC = BC$.

We call such a "triangle" $ABC$ to be degenerate. A degenerate triangle is simply a line segment.

The triangle inequality states that, if the sides of a triangle are $x$, $y$, and $z$, then the following: Note: if the inequality was not strict, it would include degenerate triangles; i.e. lines and points.

(Videos) Triangle Inequality Theorem

by Matka Kranjčević and Mr. K's Flipped Classroom

If you have 3 random sides, it is not true that you can always make a triangle. These 3 sides have to abide by certain rule, in order for you to make a triangle. Here is a video to see how this might be true:

So what is this rule that all valid triangles abide by? Watch this following video to find out.

The theorem discussed above is a variation of $\left|a+b\right|\le\left|a\right|+\left|b\right|$. We used a third side to replace $a+b$. The two versions of the formula are playing around with similar concepts.

Take a look at the triangle below. The sum of a and b is bigger than c. If you bring the blue dot down until it is on the side marked as c, the sum of a and b will equal to c.

Click here for practise questions of basic level on Triangle Inequality. Hope you get the concept right!