Problem 5

Let

a,b,c,d

be real numbers such that

a^2+b^2+c^2+d^2=1

. Determine the minimum value of

(a-b)(b-c)(c-d)(d-a)

and determine all values of

(a,b,c,d)

such that the minimum value is achieved.

Step 2 of 4: Rewrite in terms of pairwise bilinear sums

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x=ac+bd,\ y=ab+cd,\ z=ad+bc\ \Longrightarrow\ P=(x-y)(x-z),\quad x+y+z=-\tfrac{1}{2},\quad x,y,z\ge-\tfrac{1}{2}

Detailed analysis

Define $x=ac+bd$ , $y=ab+cd$ , $z=ad+bc$ . Direct expansion gives $(a-b)(b-c)(c-d)(d-a)=(x-y)(x-z)$ . Moreover $0=(a+b+c+d)^2=1+2(x+y+z)$ gives $x+y+z=-\tfrac{1}{2}$ , and $1+2x=(a+c)^2+(b+d)^2\ge0$ gives $x\ge-\tfrac{1}{2}$ (with equality when $a+c=b+d=0$ ); cyclic analogues give $y\ge-\tfrac{1}{2}$ and $z\ge-\tfrac{1}{2}$ .