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.
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.
The Erdős–Mordell inequality states that for any triangle ABC and point P inside ABC, the sum of the distances from P to the sides is less than or equal to half of the sum of the distances from P to the vertices, with equality being when the triangle is equilateral and P is the center.
Proof: Let the perpendicular feet from P to BC, CA and AB be D, E and F, respectively. Because is cyclic with PA as the diameter, or . Let the perpendicular feet from E and F to BC be M and N, respectively. Note that and . Because MN is the image of EF onto BC, . This can be written as , or . This is Mordell's Lemma.
Because of symmetry, and Adding all of these together, we get
Because , this is equivalent to
which restates the inequality.
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