In a , prove that:
Well, to start,
Now find the maximum of within the square
*Separating A and B is eluding me, and it might not even be possible. The maxima of A with respect to B seem to always be It could be possible to approximate If is sufficiently small, this will prove the hypothesis, but it's the ugly way to go about it.
Picking up where I left off, WLOG, the choice of A,B satisfying will be a stationary point in the original formula. Substitute A=M-R and B=2R and the equation becomes somewhat manageable:
So the stationary points of are precisely for . This parametric equation over M traces a line in the AB plane directly under the peaks (or valleys) of the 3-dimensional f above.
Now remember, we found the partial with respect to A, but by symmetry, the partial w.r.t. B is the exact same, simply switching A and B. Thus, We have two parametric lines running across the AB plane in the square , one exactly under the North-South ridges/valleys, and one tracing the East-West ridges/valleys. Their intersection point will be a point at which the tangent plane is perfectly horizontal. So, ~ ~ . Solving, , so .
Now, (maximum), and (no info? another way to determine?). Therefore, is the highest point on the function in . Since no legal triangle can have two right angles, WLOG, for all ,