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

In any triangle, the sum of the lengths of any two sides is greater than the third side:

Strategy: extend side $AB$ past $B$ to a point $D$ with $BD=BC$, making triangle $BCD$ isosceles so $\angle BCD=\angle BDC$. Since $\angle ACD>\angle BCD=\angle BDC$, the side opposite the larger angle in triangle $ACD$ is longer: $AD>AC$. As $AD=AB+BD=AB+BC$, this gives $AB+BC>AC$.

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 $△BCD$ isosceles. The full angle $\angle ACD$ exceeds $\angle BCD=\angle BDC$, so $AD>AC$ in $△ACD$ (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:

For any triangle $ABC$:

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. ∎