Calculateur d'inégalité triangulaire

The Triangle Inequality Theorem is one of the most fundamental statements in plane geometry: the sum of any two sides of a triangle must be strictly greater than the third side. Equivalently, no side can be longer than (or equal to) the sum of the other two. This tutorial proves the theorem, explains why "strictly greater" matters, walks through how to test any candidate set of three lengths, and shows how the same inequality generalizes to vector norms and metric spaces.

The theorem stated three ways

For any triangle with sides a, b, c, all three of the following must hold:

a + b > c
a + c > b
b + c > a

Equivalent compact statement: the longest side must be less than the sum of the other two.

The strict inequality matters. If a + b = c exactly, the "triangle" collapses to a single line segment — the three points are collinear. This degenerate case is not a triangle.

Why the theorem is true — geometric intuition

Think of building a triangle by laying three sticks end-to-end and trying to close them into a loop. Say the sticks have lengths a = 3, b = 4, c = 10.

Lay c flat on the ground. From one end, hinge stick a upward. From the other end of c, hinge stick b upward. Now try to make the free ends of a and b meet.

The longest a can stretch from its base is 3 units up (if it goes straight up). The longest b can stretch from its base is 4 units up. The two bases are 10 units apart. Even if both sticks go straight up, their free ends are 10 units apart horizontally — they cannot meet. Conclusion: no triangle exists with sides 3, 4, 10.

If we replace c = 10 with c = 6, the bases are now 6 units apart, and stick a (length 3) can reach across at most 3 units. So a + b = 7 must exceed c = 6 — and 7 > 6, so it works. The two free ends can meet at some point above the line, forming the triangle.

Formal proof — using the shortest-path principle

The shortest path between two points is the straight line connecting them. Any other path is strictly longer.

Suppose a triangle has vertices A, B, C with the side opposite each labeled a, b, c respectively. The path from A to B going directly (length c) is shorter than the path from A to C to B (length b + a). Therefore c < b + a, i.e., a + b > c.

The same argument applied to the other two pairs of vertices gives a + c > b and b + c > a.

Testing three numbers

To check whether (a, b, c) can form a triangle, you only need to test the longest side. If the longest side is less than the sum of the other two, the triangle is valid. If it equals or exceeds that sum, no triangle.

Example tests:

• (3, 4, 5): longest = 5. Other two sum to 7. 5 < 7 ✓ — valid triangle (the famous 3-4-5 right triangle).
• (5, 7, 12): longest = 12. Other two sum to 12. 12 ≥ 12 ✗ — degenerate (a straight line).
• (2, 3, 6): longest = 6. Other two sum to 5. 6 > 5 ✗ — impossible.
• (1, 1, 1): longest = 1. Other two sum to 2. 1 < 2 ✓ — valid (equilateral triangle).

Range of the third side, given two

Knowing two sides constrains the third. If a and b are given, the third side c must satisfy:

|a − b| < c < a + b

The upper bound is the triangle inequality. The lower bound is the same inequality applied to a different pairing: if c were less than or equal to |a − b|, the longer of a, b would exceed the sum c + (shorter of a, b), violating the inequality.

Example: a = 4, b = 7. Then 3 < c < 11. The third side can be any real number strictly between 3 and 11.

Why "strict inequality" matters

The boundary case a + b = c produces a "degenerate triangle" — three collinear points. Some textbooks include degenerate triangles in their definition of "triangle" (so the inequality becomes ≤). The mainstream convention requires strict inequality, and most calculators (including ours) treat the equality case as invalid.

The same theorem in vector form

The Triangle Inequality generalizes to vectors. For any two vectors u and v in any number of dimensions:

|u + v| ≤ |u| + |v|

(with equality only when u and v point in exactly the same direction, the degenerate case). This is the triangle inequality for the Euclidean norm. The same shape generalizes further to inner product spaces, normed linear spaces, and metric spaces — the inequality is one of the three defining axioms of a metric d: d(x, z) ≤ d(x, y) + d(y, z).

So the geometric triangle inequality is not just a curiosity of plane geometry — it is the defining property of "distance" in mathematics.

Common mistakes

• Only checking one of the three inequalities. All three must hold. (3, 4, 5) satisfies a + b > c but it would still need a + c > b and b + c > a — fortunately all three do. For (3, 4, 8), a + b > c fails: 3 + 4 = 7 < 8, so it's invalid. You only need to find ONE failure to rule out the triangle, but the calculator tests all three for clarity.
• Using ≥ instead of >. Strict inequality. A degenerate "triangle" with all three points collinear is not a triangle.
• Confusing "valid triangle" with "valid right triangle". Triangle inequality determines whether ANY triangle exists. To check whether the triangle is right, separately verify a² + b² = c² (Pythagorean test, where c is longest).
• Forgetting that all sides must be positive. A side length of 0 or negative cannot form a triangle regardless of what the others are.