Problem 1670

Since the expression is cyclic, we could WLOG $a=\max \{a, b, c, d\}$. Let
$$S(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)$$
Note that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \geq -\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$.

- Exactly 1 of $a-b, b-c, c-d, d-a$ is negative.

Since $a=\max \{a, b, c, d\}$, then we must have $d-a<0$. This forces $a>b>c>d$. Now, let us write
$$S(a, b, c, d)=-(a-b)(b-c)(c-d)(a-d)$$
Write $a-b=y, b-c=x, c-d=w$ for some positive reals $w, x, y>0$. Plugging to the original condition, we have
$$\begin{equation*} (d+w+x+y)^{2}+(d+w+x)^{2}+(d+w)^{2}+d^{2}-1=0 \tag{*} \end{equation*}$$
and we want to prove that $w x y(w+x+y) \leq \frac{1}{8}$. Consider the expression ( $*$ ) as a quadratic in $d$ :
$$4 d^{2}+d(6 w+4 x+2 y)+\left((w+x+y)^{2}+(w+x)^{2}+w^{2}-1\right)=0$$
Since $d$ is a real number, then the discriminant of the given equation has to be non-negative, i.e. we must have
$$\begin{aligned} 4 & \geq 4\left((w+x+y)^{2}+(w+x)^{2}+w^{2}\right)-(3 w+2 x+y)^{2} \\ & =\left(3 w^{2}+2 w y+3 y^{2}\right)+4 x(w+x+y) \\ & \geq 8 w y+4 x(w+x+y) \\ & =4(x(w+x+y)+2 w y) \end{aligned}$$
However, AM-GM gives us
$$w x y(w+x+y) \leq \frac{1}{2}\left(\frac{x(w+x+y)+2 w y}{2}\right)^{2} \leq \frac{1}{8}$$
This proves $S(a, b, c, d) \geq-\frac{1}{8}$ for any $a, b, c, d \in \mathbb{R}$ such that $a>b>c>d$. Equality holds if and only if $w=y, x(w+x+y)=2 w y$ and $w x y(w+x+y)=\frac{1}{8}$. Solving these equations gives us $w^{4}=\frac{1}{16}$ which forces $w=\frac{1}{2}$ since $w>0$. Solving for $x$ gives us $x(x+1)=\frac{1}{2}$, and we will get $x=-\frac{1}{2}+\frac{\sqrt{3}}{2}$ as $x>0$. Plugging back gives us $d=-\frac{1}{4}-\frac{\sqrt{3}}{4}$, and this gives us
$$(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}\right)$$
Thus, any cyclic permutation of the above solution will achieve the minimum equality.

- Exactly 3 of $a-b, b-c, c-d, d-a$ are negative

Since $a=\max \{a, b, c, d\}$, then $a-b$ has to be positive. So we must have $b<c<d<a$. Now, note that
$$\begin{aligned} S(a, b, c, d) & =(a-b)(b-c)(c-d)(d-a) \\ & =(a-d)(d-c)(c-b)(b-a) \\ & =S(a, d, c, b) \end{aligned}$$
Now, note that $a>d>c>b$. By the previous case, $S(a, d, c, b) \geq-\frac{1}{8}$, which implies that
$$S(a, b, c, d)=S(a, d, c, b) \geq-\frac{1}{8}$$
as well. Equality holds if and only if
$$(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right)$$
and its cyclic permutation.

The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
$$\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .$$
Cyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \rightarrow (-a,-b,-c,-d)$ ) all four entries in each of the four quadruples give four more equality cases. We then begin the proof by the following optimization:

Claim 1. In order to get the minimum value, we must have $a+b+c+d=0$.

Proof. Assume not, let $\delta=\frac{a+b+c+d}{4}$ and note that
$$(a-\delta)^{2}+(b-\delta)^{2}+(c-\delta)^{2}+(d-\delta)^{2}<a^{2}+b^{2}+c^{2}+d^{2}$$
so by shifting by $\delta$ and scaling, we get an even smaller value of $(a-b)(b-c)(c-d)(d-a)$.

The key idea is to substitute the variables
$$\begin{aligned} & x=a c+b d \\ & y=a b+c d \\ & z=a d+b c \end{aligned}$$
so that the original expression is just $(x-y)(x-z)$. We also have the conditions $x, y, z \geq-0.5$ because of:
$$2 x+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a+c)^{2}+(b+d)^{2} \geq 0 .$$
Moreover, notice that
$$0=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(x+y+z) \Longrightarrow x+y+z=\frac{-1}{2}$$
Now, we reduce to the following optimization problem.

Claim 2. Let $x, y, z \geq-0.5$ such that $x+y+z=-0.5$. Then, the minimum value of
$$(x-y)(x-z)$$
is $-1 / 8$. Moreover, the equality case occurs when $x=-1 / 4$ and $\{y, z\}=\{1 / 4,-1 / 2\}$.

Proof. We notice that
$$\begin{aligned} (x-y)(x-z)+\frac{1}{8} & =\left(2 y+z+\frac{1}{2}\right)\left(2 z+y+\frac{1}{2}\right)+\frac{1}{8} \\ & =\frac{1}{8}(4 y+4 z+1)^{2}+\left(y+\frac{1}{2}\right)\left(z+\frac{1}{2}\right) \geq 0 . \end{aligned}$$
The last inequality is true since both $y+\frac{1}{2}$ and $z+\frac{1}{2}$ are not less than zero.

The equality in the last inequality is attained when either $y+\frac{1}{2}=0$ or $z+\frac{1}{2}=0$, and $4 y+4 z+1=0$. This system of equations give $(y, z)=(1 / 4,-1 / 2)$ or $(y, z)=(-1 / 2,1 / 4)$ as the desired equality cases.

Note: We can also prove (the weakened) Claim 2 by using Lagrange Multiplier, as follows. We first prove that, in fact, $x, y, z \in[-0.5,0.5]$. This can be proved by considering that
$$-2 x+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a-c)^{2}+(b-d)^{2} \geq 0 .$$
We will prove the Claim 2, only that in this case, $x, y, z \in[-0.5,0.5]$. This is already sufficient to prove the original question. We already have the bounded domain $[-0.5,0.5]^{3}$, so the global minimum must occur somewhere. Thus, it suffices to consider two cases:

- If the global minimum lies on the boundary of $[-0.5,0.5]^{3}$. Then, one of $x, y, z$ must be -0.5 or 0.5 . By symmetry between $y$ and $z$, we split to a few more cases.
- If $x=0.5$, then $y=z=-0.5$, so $(x-y)(x-z)=1$, not the minimum.
- If $x=-0.5$, then both $y$ and $z$ must be greater or equal to $x$, so $(x-y)(x-z) \geq 0$, not the minimum.
- If $y=0.5$, then $x=z=-0.5$, so $(x-y)(x-z)=0$, not the minimum.
- If $y=-0.5$, then $z=-x$, so
$$(x-y)(x-z)=2 x(x+0.5)$$
which obtain the minimum at $x=-1 / 4$. This gives the desired equality case.

- If the global minimum lies in the interior $(-0.5,0.5)^{3}$, then we apply Lagrange multiplier:
$$\begin{aligned} \frac{\partial}{\partial x}(x-y)(x-z)=\lambda \frac{\partial}{\partial x}(x+y+z) & \Longrightarrow 2 x-y-z=\lambda . \\ \frac{\partial}{\partial y}(x-y)(x-z)=\lambda \frac{\partial}{\partial y}(x+y+z) & \Longrightarrow z-x=\lambda . \\ \frac{\partial}{\partial z}(x-y)(x-z)=\lambda \frac{\partial}{\partial z}(x+y+z) & \Longrightarrow y-x=\lambda . \end{aligned}$$
Adding the last two equations gives $\lambda=0$, or $x=y=z$. This gives $(x-y)(x-z)=0$, not the minimum.

Having exhausted all cases, we are done.