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The parameters most commonly appearing in triangle inequalities are: the side lengths a, b, and c;; the semiperimeter s = (a + b + c) / 2 (half the perimeter p);; the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
The right side is the area of triangle ABC, but on the left side, r + z is at least the height of the triangle; consequently, the left side cannot be smaller than the right side. Now reflect P on the angle bisector at C. We find that cr ≥ ay + bx for P's reflection. Similarly, bq ≥ az + cx and ap ≥ bz + cy. We solve these inequalities for ...
In spherical geometry, the shortest distance between two points is an arc of a great circle, but the triangle inequality holds provided the restriction is made that the distance between two points on a sphere is the length of a minor spherical line segment (that is, one with central angle in [0, π]) with those endpoints.
to construct a triangle whose side lengths are the products of sides of the given quadrilateral, and applying the triangle inequality to this triangle. [6] One can also view the points as belonging to the complex projective line, express the inequality in the form that the absolute values of two cross-ratios of the points sum to at least one ...
Azuma's inequality; Bennett's inequality, an upper bound on the probability that the sum of independent random variables deviates from its expected value by more than any specified amount
The triangle inequality states that the length of two sides of any triangle, added together, will be equal to or greater than the length of the third side. In simplest terms, the Kantorovich inequality translates the basic idea of the triangle inequality into the terms and notational conventions of linear programming .
The Ruzsa sum triangle inequality is a corollary of the Plünnecke-Ruzsa inequality (which is in turn proved using the ordinary Ruzsa triangle inequality). Theorem (Ruzsa sum triangle inequality) — If A {\displaystyle A} , B {\displaystyle B} , and C {\displaystyle C} are finite subsets of an abelian group, then
The reverse inequality follows from the same argument as the standard Minkowski, but uses that Holder's inequality is also reversed in this range. Using the Reverse Minkowski, we may prove that power means with p ≤ 1 , {\textstyle p\leq 1,} such as the harmonic mean and the geometric mean are concave.