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]}