A problem in elementary geometry Let us have a triangle ABC  in the Cartesian plane and consider the following transformation of this triangle: 
On  the ray  AB starting at A, select a point B' so that  so that |AB'|=|AC|. Likewise,
 on the ray BC starting at B, select a point C' so that  BC' so that |BC'|=|AB|,
 and on the ray CA starting at C select a point A' so that |BA'|=|BC|.
 As a result, a new triangle A'B'C' is obtained.
Prove (or disprove)  that the perimeter of a new triangle A'B'C'  does not exceed the perimeter of ABC.
Remark. The question has been stated by Ahmet Yaşar Özban, Atilim University in regard with his research on iterative methods.
 A: Let O be the point of intersection of the angle bisectors of the initial triangle ABC (the incenter). Then the length of OB' is equal to the length of OC, |OC'|=|OA| and |OA'|=|OB|. Now the statement of the problem will follow from the following fact: given point O and three length x,y,z, consider all triangles EDF that satisfy |OE|=x, |OD|=y, |OF|=z; among those the triangle with the greatest perimeter is unique (up to congruence) and it is the only triangle in the family for which O is the incenter. 
Consider the maximum point, clearly it exists, fix two of the rays, say, OE and OD, and let the third point F move along the circle centered at O. If the triangle EDF maximizes the perimeter, then this circle is tangent to the ellipse with foci at E and D, then by the property of the tangent of an ellipse, OF is the bisector of the angle EFD. This implies that for the triangle of maximal perimeter O is the incenter. 
The last remark is that there is only one triangle with fixed distances from the incenter to the vertices, it follows for example by applying the sine theorem to small triangles EOF, FOD and DOE.
