The Triangle Inequality

The detour is never shorter

In any triangle, pick two of the corners. Walking straight from one to the other covers the side between them; call it c. Going round by way of the third corner covers the other two sides, a and b. A straight line is the shortest path between two points, so the detour is never shorter: $a+b\ge c$a + b ≥ c.

The same holds for every side in turn: $b+c\ge a$b + c ≥ a and $a+c\ge b$a + c ≥ b. Each side of a triangle is at most the sum of the other two.

For a real triangle, with its three corners not on one line, the inequality is strict: a + b > c. Equality, a + b = c, happens only when the third corner lies on the side c itself, so the triangle is flat: three points on a line.

Walk straight from one corner to another along c, or go round by way of the third corner along a and b.

Squeezed from both ends

Take two sides 5 and 4 and change the angle between them. As the angle opens toward 180°, the third side c grows toward 5 + 4 = 9 but stays below it while there is a triangle at all. As the angle closes toward 0°, c shrinks toward 5 − 4 = 1 and stays above it.

The lower limit is the triangle inequality again, for a different side: $5\le 4+c$5 ≤ 4 + c, so $c\ge 1$c ≥ 1. Together, for any triangle, the third side is more than the difference of the other two and less than their sum, and it reaches neither unless the triangle is flat.

Sides 5 and 4 with the angle between them opened to 150°. The third side is about 8.7: close to 5 + 4 = 9, and still under it.

The same two sides with the angle closed to 30°. The third side is about 2.5, still more than 5 − 4 = 1.

Testing three lengths

Three lengths make a triangle exactly when each is less than the sum of the other two. It is enough to test the longest against the sum of the two shorter ones, because the longest side plus either shorter side is already more than the remaining side.

Sides 5, 9 and 15: 5 + 9 = 14, which is less than 15. The two shorter sides laid end to end do not reach across the longest, so no triangle forms. Sides 6, 8 and 10: 6 + 8 = 14, which is more than 10, so they make a triangle. Sides 2, 3 and 5: 2 + 3 = 5, exactly the longest, so the three lie flat in a line and enclose no area.

With two sides 5 and 9, a third side x must be less than 5 + 9 = 14, and more than 9 − 5 = 4, since 9 < 5 + x. The largest whole number it can be is 13; at 14 the three lie flat.

The two shorter sides laid end to end: 5 and then 9 reach 14, one short of the longest side, 15.

With modulus bars

For numbers, |a| is the distance from a to 0 on the number line, and the triangle inequality reads $|a+b|\le |a|+|b|$|a + b| ≤ |a| + |b| for any real numbers a and b.

The proof: every number lies within its own distance of 0, so $-|a|\le a\le |a|$−|a| ≤ a ≤ |a| and $-|b|\le b\le |b|$−|b| ≤ b ≤ |b|. Adding the two gives $-\left(|a|+|b|\right)\le a+b\le |a|+|b|$−(|a| + |b|) ≤ a + b ≤ |a| + |b|. So a + b lies no farther than |a| + |b| from 0 on either side, and that is $|a+b|\le |a|+|b|$|a + b| ≤ |a| + |b|.

Equality holds exactly when a and b have the same sign, or one of them is 0. Then both moves along the number line go the same way and the distances add: |3 + 4| = 7 = |3| + |4|, and |−2 + (−6)| = 8 = |−2| + |−6|. When one is positive and the other negative, the second move goes back over part of the first, so a + b lies strictly between them and |a + b| is less than the larger of |a| and |b|: |5 + (−3)| = 2, while |5| + |−3| = 8.

a = 5 and then b = −3. The two moves cover 5 + 3 = 8 units, but the second goes back, and a + b = 2 ends only 2 from 0.

a = 3 and then b = 4, the same sign. Both moves go the same way, so a + b ends 3 + 4 = 7 from 0, and the inequality is an equality.

Vectors and complex numbers

For vectors the bars mean length, and the statement is the triangle itself. Placed tip to tail, u and v and their sum u + v are the three sides of a triangle, so $|u+v|\le |u|+|v|$|u + v| ≤ |u| + |v|, with equality only when u and v point the same way or one of them is zero.

For example, (3, 4) has length 5 and (5, 12) has length 13. Their sum (8, 16) has length $\sqrt{8^2+16^2}\approx 17.89$√(8² + 16²) ≈ 17.89, which is less than 5 + 13 = 18. A complex number is a point of the plane and its modulus is its distance from 0, so complex numbers obey the same inequality, $|z+w|\le |z|+|w|$|z + w| ≤ |z| + |w|.

The usual mistakes

Accepting a sum equal to the longest side. Sides 2, 3 and 5 lie flat; a triangle needs the sum of the two shorter sides to be strictly greater than the longest.

Testing the wrong pair. Compare the longest side with the sum of the other two: 5 + 15 > 9 says nothing about sides 5, 9 and 15.

Forgetting the lower limit. With sides 5 and 9 the third side must be more than 4 as well as less than 14.

Expecting |a + b| = |a| + |b| every time. With opposite signs the left side is smaller.

A depot and planter frames

In the applications below, a depot on the way from a warehouse to a store bounds the length of the direct road from above and below, and a company rule about detours becomes a second inequality. Then a workshop tests which choices of stock rods weld into triangular frames, and finds the range of a third rod.