The Triangle Inequality says that the sum of the lengths of any two sides of a nondegenerate triangle is greater than the length of the third side. This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric space in analysis.
The Pythagorean Inequality is a generalization of the Pythagorean Theorem. The Theorem states that in a right triangle with sides of length we have . The Inequality extends this to obtuse and acute triangles. The inequality says:
For an acute triangle with sides of length , . For an obtuse triangle with sides , .
The Isoperimetric Inequality states that if a figure in the plane has area and perimeter , then . This means that given a perimeter for a plane figure, the circle has the largest area. Conversely, of all plane figures with area , the circle has the least perimeter.
- In , .
Alternatively, we may use a method that can be called "perturbation". If we let all the angles be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if , then . This inequality reduces to , which is equivalent to . Since this is always true for , this inequality is true. Therefore, the maximum value of this expression is when , which gives us the value .
Similarly, in , .
Euler's inequality states that with equality when is equailateral, where and denote the circumradius and inradius of triangle , respectively.
Proof: The distance from the circumcenter and incenter of a triangle can be expressed as , meaning or equivalently with equality if and only if the incenter equals the circumcenter, namely the triangle is equilateral.
Ptolemy's inequality states that for any quadrilateral , with equality when quadrilateral is cyclic.
First Proof: Let P be the point such that . By SAS we also have that . By the triangle inequality, . calculating the lengths, we obtain an equivalent statement: . Multiplying by we get the desired result with equality when P is on DC. This happens when . But so , or quadrilateral is cyclic.
Second Proof (using inversion): Let the inversion map B,C and D to B',C' and D' respectively. We then have By the triangle inequality, we have By multiplying on both sides we get the desired result with equality when is collinear, implying either ABCD is cyclic or collinear.
The Erdős–Mordell inequality states that if lies in then where are the foot of the altitudes from to and
Proof: First, we prove a lemma.
Proof of Lemma: Let and be the projections of and onto line Note that is cyclic with diameter By ELOS, Since is cyclic, we have that and are supplementary. Since is a line, This means that Similarly, So the problem is reduced to proving that but this is obvious by the Pythagorean Theorem.
Now the rest of the problem is straightforward. We know that Adding these cyclically implies By AM-GM, with equality when ABC is equilateral and P is the center of it. Cagegory:Geometric Inequalities This article is a stub. Help us out by .