Suppose we have the following identity: #(px + (1-p)y)^2 = Ax^2 + Bxy + Cy^2.# Find the minimum of #max(A,B,C)# over #0 leq p leq 1#?

1 Answer
Jun 9, 2017

The minimum value of #max(A,B,C)=4/9#.

Explanation:

#(px+(1-p)y)^2=Ax^2+Bxy+Cy^2#

Expand the right-hand side:
#p^2x^2+2p(1-p)xy+(1-p)^2y^2=Ax^2+Bxy+Cy^2#

Equate coefficients:
#{(p^2=A),(2p(1-p)=B),((1-p)^2=C):}#

Let us find when any of the two variables intersect (as this will determine when #max(A,B,C)# will output a different variable)
#A=B#
#p^2=2p(1-p)#
#3p^2-2p=0#
#p=0,2/3# (remembering that #0≤p≤1#)

#A=C#
#p^2=(1-p)^2#
#p=1/2#

#B=C#
#2p(1-p)=(1-p)^2#
#(1-p)(1-p-2p)=0#
#p=1/3,1#

Substituting arbitrary values in the ranges #0≤p≤1/3#, #1/3≤p≤1/2#, #1/2≤p≤2/3#, and #2/3≤p≤1# shows that #max(A,B,C)=C# for #0≤p≤1/3#, #max(A,B,C)=B# for #1/3≤p≤2/3#, and #max(A,B,C)=A# for #2/3≤p≤1#.

When #0≤p≤1/3#, #C# is minimum when #p=1/3# with value #4/9#. When #1/3≤p≤2/3#, #B# is minimum when #p=1/3# or #p=2/3# with value #4/9#. When #2/3≤p≤1#, #A# is minimum when #p=2/3# with value #4/9#.

Thus, the minimum value of #max(A,B,C)=4/9#.

This can be shown with a graph:
graph{(y-x^2)(y-(1-x)^2)(y-2x(1-x))=0 [-0.1, 1.1, -0.1, 1.1]}