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 3 of 4: Algebraic identity for the lower bound -1/8

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(x-y)(x-z)+\frac{1}{8}=\frac{1}{8}(4y+4z+1)^2+\left(y+\frac{1}{2}\right)\left(z+\frac{1}{2}\right)\ge0

Detailed analysis

Eliminating $x=-\tfrac{1}{2}-y-z$ gives $(x-y)(x-z)+\tfrac{1}{8}=(2y+z+\tfrac{1}{2})(y+2z+\tfrac{1}{2})+\tfrac{1}{8}=\tfrac{1}{8}(4y+4z+1)^2+(y+\tfrac{1}{2})(z+\tfrac{1}{2})$ . Since $y+\tfrac{1}{2}\ge0$ and $z+\tfrac{1}{2}\ge0$ , the right-hand side is non-negative, so $(x-y)(x-z)\ge-\tfrac{1}{8}$ . Equality holds if and only if $4y+4z+1=0$ and $(y+\tfrac{1}{2})(z+\tfrac{1}{2})=0$ , i.e. $x=-\tfrac{1}{4}$ and $\{y,z\}=\{-\tfrac{1}{2},\tfrac{1}{4}\}$ .