Here is a problem that I've proposed which I believe to be true but I haven't verified it .
Take any triangle ABC and three concurrent lines emanating from each vertex A,B,C which extend to the sides x,y,z respectively .Prove that the area of triangle AZY + triangle ZBX + triangle YXC >=3(area of XZY). Equality is reached when the points x,y and z are the mid points of the sides of the triangle .
1. "RE: A Geometry Inequality"
In response to message #0
What you have to do is to consider the barycentric coordinates of the point of concurrency of the three lines and how in terms of those coordinates they lines split the opposite sides of ΔABC: