The Triangle Inequality

A proof that the sum of any two sides of a triangle is greater than the third side, using an isosceles-triangle extension argument.

Keywords: triangle inequality, side lengths, isosceles triangle

Prerequisites: angle-side-inequality · Difficulty: beginner

The triangle inequality states that in any triangle, the sum of the lengths of any two sides is greater than the length of the third side:

AB + BC > AC, \quad AB + AC > BC, \quad AC + BC > AB
    

This is a fundamental fact — it tells us exactly which triples of positive numbers can be side lengths of a triangle.

Proof Strategy

We extend side $AB$ past $B$ to a point $D$ such that $BD = BC$. This creates an isosceles triangle $BCD$ where $\angle BCD = \angle BDC$. Since $\angle ACD > \angle BCD = \angle BDC$, the side opposite the larger angle in triangle $ACD$ is longer: $AD > AC$.

But $AD = AB + BD = AB + BC$, so $AB + BC > AC$.

The Proof

Step 1: The Triangle

We want to prove that $AB + BC > AC$ for any triangle $ABC$.

Step 2: Triangle Inequality

Claim: For any triangle $ABC$, the sum of any two sides exceeds the third: $$AB + BC > AC.$$ By symmetry, the same holds for the other two pairs.

Proof sketch. Extend $AB$ past $B$ to $D$ with $BD = BC$, making $\triangle BCD$ isosceles. The full angle $\angle ACD$ exceeds $\angle BCD = \angle BDC$, so $AD > AC$ in $\triangle ACD$ (longer side opposite larger angle). Since $AD = AB + BD = AB + BC$, we get $AB + BC > AC$.

Step 3: Extend Side $AB$ past $B$

Extend side $AB$ past $B$ to a point $D$ such that $BD = BC$. This makes triangle $BCD$ isosceles.

Step 4: Isosceles Triangle Gives Equal Angles

Since $BD = BC$, triangle $BCD$ is isosceles. Therefore its base angles are equal: $\angle BCD = \angle BDC$.

Step 5: Conclude $AB + BC > AC$

Now look at the full angle $\angle ACD$. It is larger than $\angle BCD$ (since $\angle ACD$ contains $\angle BCD$ plus extra).

So $\angle ACD > \angle BCD = \angle BDC = \angle ADC$.

In triangle $ACD$, the side opposite the larger angle is longer. Since $\angle ACD > \angle ADC$, we get $AD > AC$.

But $AD = AB + BD = AB + BC$. Therefore:

AB + BC > AC
    

Conclusion

For any triangle $ABC$:

AB + BC > AC
    

By symmetry, this holds for every pair of sides. The triangle inequality is a necessary and sufficient condition: three positive lengths form a triangle if and only if every pair sums to more than the third.