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 1 of 4: Force zero sum by shifting and rescaling

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\text{At the minimum, }a+b+c+d=0

Detailed analysis

Since $4$ -tuples attaining negative values of $P=(a-b)(b-c)(c-d)(d-a)$ clearly exist, the minimum of $P$ is strictly negative. If $s=(a+b+c+d)/4\ne0$ , shifting each variable by $-s$ leaves all differences (hence $P$ ) unchanged while strictly decreasing $a^2+b^2+c^2+d^2$ to $1-4s^2<1$ ; rescaling the shifted tuple by $(1-4s^2)^{-1/2}>1$ restores the unit sum of squares and multiplies the negative number $P$ by $(1-4s^2)^{-2}>1$ , making it strictly more negative. Hence any minimizer satisfies $a+b+c+d=0$ .