The comparison of lengths is more basic than comparison of other geometric quantities (such as angles, areas and volumes). A geometric inequality that involves only the lengths is called a distance inequality.
Some simple axioms and theorems on inequalities in Euclidean geometry are usually the starting point to solve the problem of distance inequality, in which most frequently used tools are:
The shortest line connecting point A with point B is the segment AB.
The direct corollary of proposition 1 is as follows:
For arbitrary three points A, B and C (lying in the same plane), we have , the equality holds iff the three points are collinear.
In most literature, any symbol of a geometric object also denotes its quantity according to the context.
In a triangle, the longer side has the larger opposite angle. And, conversely, the longer angle has the longer opposite side.
The median of a triangle on a side is shorter than the half-sum of the other two sides.
If a convex polygon is within another one, then the outside convex polygon’s perimeter is larger.
Any segment in a convex polygon is either less than the longest side or the longest diagonal of the convex polygon.
Here, is a classic example:
Let be the sides of a . Prove that .
By the triangle inequality, yields
Adding up the above three inequalities, we get the required inequality.
Let AB be the longest side of , and P a point in the triangle, prove that .
Let D be the intersection point of CP and AB. (Note P is in the interior of the triangle ABC). Then, or is not acute. Without loss of generality, we assume that is not acute. Applying proposition 3 to , we obtain . Therefore, ….call this as relationship “a”.
Furthermore, applying triangle inequality to , we have …call this as relationship “b”.
Combining “a” and “b”, we obtain the required inequality immediately. QED.
Remarks: (1) If AB is not the longest, then the conclusion may not be true. (2) If point P on the plane of regular triangle ABC, P is not on the circumcircle of the triangle, then the sum of any two of PA, PB, and PC is longer than the remaining one. That is, PA, PB and PC consist of a triangle’s three sides.
Prove that a closed polygonal line with perimeter 1 can be put inside a circle with radius 0.25.
Geometric Inequalities, Vol 12, Gangsong Leng, translated by Yongming Liu.
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