Question

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`?

Answer 1

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]}