# When $a^2 + b^2 + c^2 + d^2 = 1$ and the expression $(a - b)(b - c)(c - d)(d - a)$ reaches its minimum value, determine the value of product $abcd$.

Consider real number $$a, b, c, d$$ such that $$a^2 + b^2 + c^2 + d^2 = 1$$. When the expression $$(a - b)(b - c)(c - d)(d - a)$$ reaches its minimum value, determine the value of product $$abcd$$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)

By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]

So I took the mock exam which was organised by our school today... It didn't go well. ｡ﾟ( ﾟஇ‸இﾟ)ﾟ｡ That's why this question is here.

Here's what I'd attempted in the official amount of time that was given, (so no external thoughts, I still have more subjects that need to be taken care of, sorry~)

(I lied, there are some afterthoughts sprinkled in here.)

First of all, \begin{aligned} (a - b)(b - c)(c - d)(d - a) &= [(ac + bd) - (ab + cd)][(ac + bd) - (bc + da)]\\ &= (ac + bd)^2 - (ab + bc + cd + da)(ac + bd)\\ &+ (ab + cd)(bc + da) \end{aligned}

(From this point on, it's all what I've thought of after the official time period has ended.)

Let $$ac + bd = x$$. The minimum value of the quadratic function $$x^2 - (ab + bc + cd + da)x + (ab + cd)(bc + da)$$ is $$\dfrac{-\Delta}{4} = \dfrac{4(ab + cd)(bc + da) - (ab + bc + cd + da)^2}{4} = -\left[\dfrac{(a - c)(b - d)}{2}\right]^2$$.

The equal sign occurs when $$x = \dfrac{ab + bc + cd + da}{2} \iff 2(ac + bd) = ab + bc + cd + da \ (1)$$. Now, we just need to find the maximum value of $$\left[\dfrac{(a - c)(b - d)}{2}\right]^2$$.

(∩ᗒ.ᗕ)⊃━☆ﾟ.❉*, $$\left[\dfrac{(a - c)(b - d)}{2}\right]^2 \le \dfrac{(a - c)^2 + (b - d)^2}{8} \le \dfrac{a^2 + b^2 + c^2 + d^2}{4} = \dfrac{1}{4}$$.

The equal sign occurs when $$a = b = -c = -d \ (2)$$. Combining $$(1)$$ and $$(2)$$, we have that \left\{ \begin{aligned} 2(ac + bd) &= ab + bc + cd + da\\ a = b &= -c = -d \end{aligned} \right. \implies a = b = c = d = 0

This feels wrong, and of course, it is. Let's elaborate on $$(1)$$ further, in that, \begin{aligned} 2(ac + bd) = ab + bc + cd + da &\iff 2[(ac + bd) - (bc + da)] = ab - bc + cd - da\\ &\iff 2(a - b)(c - d) = (a - c)(b - d) \end{aligned}

Now, does this mean anything? It is with deep regret that I have to inform you that, I don't freaking know.

Hmmm~ this doesn't seem to work. Let's try something different, for example, assuming that $$a \ge b \ge c \ge d$$, we have that \begin{aligned} (a - b)(b - c)(c - d)(d - a) &\ge \dfrac{[(a - b) + (b - c) + (c - d)]^3(d - a)}{27} = \dfrac{-(a - d)^4}{27} \end{aligned}

The equal sign occurs when a - b = b - c = c - d \iff \left\{ \begin{aligned} b = \dfrac{d + 2a}{3}\\ c = \dfrac{2d + a}{3} \end{aligned} \right. \ (3).

(From this point on, it's all what I've thought of after the official time period has ended, the sequel.)

Plugging the above results into $$a^2 + b^2 + c^2 + d^2 = 1$$, we have that $$7d^2 + 4da + 7a^2 = \dfrac{9}{2}$$, and since $$(7d^2 + 4da + 7a^2) - \dfrac{5(d - a)^2}{2} = \dfrac{9(d + a)^2}{2} \ge 0, \forall d, a \in \mathbb R$$, it means that $$\dfrac{5(d - a)^2}{2} \le \dfrac{9}{2} \iff (d - a)^2 \le \dfrac{9}{5} \iff\dfrac{-(a - d)^4}{27} \ge -\dfrac{3}{25}$$

The equal sign occurs when $$d = -a \ (4)$$.

Combining $$(3)$$ and $$(4)$$, we have that \left[ \begin{aligned} (a; b; c; d) &= \left(\dfrac{3\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; -\dfrac{3\sqrt{5}}{10}\right)\\ (a; b; c; d) &= \left(-\dfrac{3\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; \dfrac{3\sqrt{5}}{10}\right) \end{aligned} \right. \implies abcd = \dfrac{9}{400}

According to WolframAlpha, the minimum value of the above expression is $$-\dfrac{1}{8}$$, happening when $$(a; b; c; d) = \left(-\dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 - 1}{4}; -\dfrac{\sqrt 3 - 1}{4}\right) \implies abcd = \dfrac{1}{64}$$

So my friend was right after all...（；¬＿¬) How am I supposed to do this in 90 minutes?

All anger aside, that's all for now. This took way more time to write down than it needed to. Anyhow, have a wonderful tomorrow, everyone~

By the way, the choices were $$-\dfrac{1}{64}$$; $$-\dfrac{1}{8}$$; $$\dfrac{1}{8}$$ and $$\dfrac{1}{64}$$.

• The $a,b,c,d$ values from WolframAlpha are roots of $4 (4x^2-1)^2 = 3$, if it helps. (But I don't know why yet.) Then the corresponding product $abcd$ matches what Vieta's formula gives. Commented Jun 3, 2022 at 15:06
• If I keep your assumption that $a\ge b\ge c\ge d$, drop $a-b=b-c=c-d$, and instead relax it to just "symmetric" assumptions $a\ge b > 0$, $c=-b$, $d=-a$, then \begin{align*} (a-b)(b-c)(c-d)(d-a) &= (a-b)\cdot 2b \cdot (a-b)\cdot (-2a)\\ &= -4ab(a^2 - 2ab + b^2)\\ &= -2ab(1-4ab) \end{align*} Minimum is when $ab =\frac{1}{8}$, which can be achieved because $0\le ab \le \frac 14$. Corresponding $abcd = (ab)^2 = \frac1{64}$. I am not sure if the "symmetric" assumptions are valid though. Commented Jun 3, 2022 at 15:27
• Just an observation $(a; b; c; d) = \left(\dfrac{\sin (13\pi/12)}{\sqrt2}; \dfrac{\sin (5\pi/12)}{\sqrt2}; \dfrac{\cos (5\pi/12)}{\sqrt2}; \dfrac{\cos (13\pi/12)}{\sqrt2}\right)$
– user1012971
Commented Jun 3, 2022 at 15:39
• The objective function $(a-b)(b-c)(c-d)(d-a)$ is invariant under both even permutations and simultaneous negation of all variables. If the minimum is unique for $a\ge b\ge c \ge d$, then we must have $a = -d$ and $b = -c$. Commented Jun 3, 2022 at 16:59
• The choices were $-\dfrac{1}{64}$; $-\dfrac{1}{8}$; $\dfrac{1}{8}$ and $\dfrac{1}{64}$. Since $a^2b^2c^2d^2 \le \left(\frac{a^2+b^2+c^2+d^2}{4}\right)^4 = \frac{1}{4^4}$, we have $|abcd| \le \frac{1}{16}$. So the remaining choices are $-1/64$ and $1/64$. Is there easy way to exclude $-1/64$? Commented Jun 6, 2022 at 0:20

Let $$f(a,b,c,d) := (a-b)(b-c)(c-d)(d-a)$$.

We have $$f_{\min} < 0$$, by noting that $$f < 0$$ when $$a > b > c > d$$.

Since the constraint is symmetric and $$f$$ is cyclic, WLOG, assume that $$a = \max(a, b, c, d)$$. At minimum, we have $$a - b > 0$$ and $$d - a < 0$$, so $$(b - c)(c - d) > 0$$ which results in $$a > b > c > d$$ or $$a > d > c > b$$. Note also that $$f(a, b, c, d) = f(a, d, c, b)$$. Thus, we only need to consider the case when $$a > b > c > d$$.

Let $$x = a - b, \, y = b - c, \, z = c - d$$. Then $$x, y, z > 0$$ and $$c = d + z, \, b = d + y + z, \, a = d + x + y + z$$. We have $$f = - xyz (x + y + z).$$

From $$a^2+b^2+c^2+d^2 = 1$$, we have $$4d^2 + (2x + 4y + 6z)d + x^2 + 2xy + 2xz + 2y^2 + 4yz + 3z^2 - 1 = 0.$$ This is a quadratic equation in $$d$$ which has a non-negative discriminant i.e. $$-12x^2 - 16xy - 8xz - 16y^2 - 16yz - 12z^2 + 16 \ge 0$$ which is written as (noting $$f = -xz \cdot y(x + y + z)$$) $$-12(z + x)^2 + 16xz + 16 \ge 16y (x + y + z). \tag{1}$$ Using $$zx \le (z+x)^2/4$$ and (1), we have $$16 - 8(x + z)^2 \ge 16y (x + y + z). \tag{2}$$ From (2), we have $$16 > 8(x + z)^2$$.

Using (2) and $$zx \le (z+x)^2/4$$ and AM-GM, we have \begin{align*} f &= -xyz(x + y + z)\\ &\ge - xz\cdot (1 - (x+z)^2/2)\\ &\ge - \frac{(x + z)^2}{4} \cdot (1 - (x + z)^2/2) \\ &= - \frac{1}{8}(x + z)^2 (2 - (x + z)^2)\\ &\ge - \frac18 \cdot \frac{[(x + z)^2 + 2 - (x + z)^2]^2}{4}\\ &= - \frac18 \end{align*} with equality if and only if $$x = z$$ and $$(x + z)^2 = 1$$ and $$y(x + y + z) = 1 - (x + z)^2/2$$ i.e. $$x = z = 1/2$$ and $$y = \frac{\sqrt 3 - 1}{2}$$. It is not difficult to get $$a = \frac{\sqrt 3 + 1}{4}, b = \frac{\sqrt 3 - 1}{4}$$ and $$c = -b, d = -a$$. Thus, $$abcd = 1/64$$.

I suppose there must be more simple and short solution, but now I post this.

$$A=(a-b) (b-c) (c-d) (a-d)=\max$$ $$B=a^2+b^2+c^2+d^2=1$$ $$C=a b c d=\text{to find}$$

$$A$$ can be greater than 0, so we consider only case $$A>0$$. For $$A$$ to be greater than 0 no one of multipliers can't be 0.

If we change all signs of $$a, b, c, d$$ then $$A$$ and $$B$$ don't change, then WLOG we can take $$a-b>0 \Rightarrow a>b$$

$$A=-(a-b)(c-b)(a-d)(c-d)$$ $$A$$ is symmetric with respect to $$b$$ and $$d$$. $$b$$ cannot be equal $$d$$ otherwise $$A=-(a-b)^2(c-b)^2\leq0$$. Then WLOG $$b>d$$

$$a>b>d$$, then $$(a-b)(a-d)>0$$ then $$(b-c)(c-d)$$ must be greater than 0. Taking into account $$b>d$$, we get $$a>b>c>d$$.

Let $$e=\frac{b+c}2$$, $$x=b-e=e-c$$, $$y=a-b$$, $$z=c-d$$. Then $$A=2x y z (2x+y+z),B=(e-x)^2+(e+x)^2+(e+x+y)^2+(e-x-z)^2$$

If we multiply all $$a,b,c,d$$ by factor $$k$$, $$A$$ will become $$k^4A$$ and $$B$$ will become $$k^2A$$. Let consider two sets of $$a,b,c,d$$ satisfying $$B=1$$: one set is optimal giving $$A=A_{\max}$$ and another set is suboptimal giving $$0. If we factor suboptimal set in such way that $$A$$ will become $$A_\max$$ then $$B$$ will become greater than 1. Then optimal set of $$a,b,c,d$$ is also solution of following problem: $$A=A_{\max}$$, $$B=\min$$.

Let consider this problem in terms of $$x,y,z,e$$: $$A=2x y z (2x+y+z)=A_{\max },B=(e-x)^2+(e+x)^2+(e+x+y)^2+(e-x-z)^2$$

Using Lagrange method: $$\Phi =B+\lambda \left(A-A_{\max }\right)=(e-x)^2+(e+x)^2+(e+x+y)^2+(e-x-z)^2+\lambda \left(2x y z (2x+y+z)-A_{\max }\right)$$ $$\frac{\partial \Phi }{\partial e}=2 (4 e+y-z)=0\Rightarrow e=\frac{z-y}{4}\Rightarrow B=\frac{16 x^2+8 x y+3 y^2+8 x z+2 y z+3 z^2}{4}$$ $$\Phi =\frac{16 x^2+8 x y+3 y^2+8 x z+2 y z+3 z^2}{4}+\lambda \left(2x y z (2x+y+z)-A_{\max }\right)$$ $$\frac{\partial \Phi }{\partial y}=\frac{4 x+3 y+z+8 x^2 z \lambda +8 x y z \lambda +4 x z^2 \lambda }{2} =0\Rightarrow \lambda =\frac{-4 x-3 y-z}{4 x z (2 x+2 y+z)}$$ $$\frac{\partial \Phi }{\partial z}=\frac{4 x+y+3 z+8 x^2 y \lambda +4 x y^2 \lambda +8 x y z \lambda }{2} =0\Rightarrow \lambda =\frac{-4 x-y-3 z}{4 x y (2 x+y+2 z)}$$ $$\lambda =\frac{-4 x-3 y-z}{4 x z (2 x+2 y+z)}=\frac{-4 x-y-3 z}{4 x y (2 x+y+2 z)}\Rightarrow \frac{-4 x-3 y-z}{4 x z (2 x+2 y+z)}-\frac{-4 x-y-3 z}{4 x y (2 x+y+2 z)}=0\Rightarrow\\ -\frac{(y-z) \left(8 x^2+10 x y+3 y^2+10 x z+8 y z+3 z^2\right)}{4 x y z (2 x+2 y+z) (2 x+y+2 z)}=0$$

$$x,y,z>0$$ then minimum point must satisfy $$y-z=0\Rightarrow y=z$$. Then $$A=4x z^2(x+z),B=4 x^2+4 x z+2 z^2$$ $$\Phi =4 x^2+4 x z+2 z^2+\lambda \left(4x z^2(x+z)-A_{\max }\right)$$ $$\frac{\partial \Phi }{\partial z}=4 \left(x+z+2 x^2 z \lambda +3 x z^2 \lambda \right)=0\Rightarrow \lambda =-\frac{x+z}{x z (2 x+3 z)}$$ $$\frac{\partial \Phi }{\partial x}=4 (2 x+z) \left(1+z^2 \lambda \right)=0\Rightarrow \lambda =-\frac{1}{z^2}$$ $$\lambda =-\frac{x+z}{x z (2 x+3 z)}=-\frac{1}{z^2}\Rightarrow \frac{x+z}{x z (2 x+3 z)}-\frac{1}{z^2}=0\Rightarrow\\ \frac{-2 x^2-2 x z+z^2}{x z^2 (2 x+3 z)}=0\Rightarrow z=x\left(1+\sqrt{3}\right)$$

Then $$A=4x z^2(x+z)=2\left(1+\sqrt{3}\right)^4 x^4,B=4 \left(1+\sqrt{3}\right)^2 x^2$$

$$B=1\Rightarrow x^2=\frac{1}{4\left(1+\sqrt{3}\right)^2}\Rightarrow x=\frac{1}{2\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}-1}{4}$$ $$z=x\left(1+\sqrt{3}\right)=\frac{1}{2},y=z=\frac{1}{2}$$ $$\frac{b+c}{2}=e=\frac{y-z}{4}=0$$ $$b=e+x=\frac{\sqrt{3}-1}{4},c=e-x=-\frac{\sqrt{3}-1}{4},a=b+y=\frac{\sqrt{3}+1}{4},d=c-z=-\frac{\sqrt{3}+1}{4}$$ $$a b c d=\left(\frac{\sqrt{3}-1}{4}\right)^2\left(\frac{\sqrt{3}+1}{4}\right)^2=\frac{1}{64}$$

• @MartinR: That caught me too, but then I noticed that he writes $A = (a-b)(b-c)(c-d)(a-d)$. A subtle change. Commented Jun 3, 2022 at 20:00

My second solution:

Let $$x = (a + b)/2, y = (a - b)/2, z = (c + d)/2, w = (c - d)/2$$ (correspondingly, $$a = x + y, b = x - y, c = z + w, d = z - w$$).

We have $$x^2 + y^2 + z^2 + w^2 = (a^2 + b^2 + c^2 + d^2)/2 = 1/2$$.

We have $$f(a,b,c,d) : =(a-b)(b-c)(c-d)(d-a) = 4yw(w+y)^2-4yw(x-z)^2$$ and $$g(a,b,c,d) := abcd = -w^2x^2 + w^2y^2 + x^2z^2 - y^2z^2.$$

If $$yw \le 0$$, we have \begin{align*} f &\ge 4yw(w + y)^2\\ & = - 4 \cdot |yw| \cdot (w^2 + y^2 - 2|yw|)\\ &\ge -2\cdot \frac{(2|yw| + w^2 + y^2 - 2|yw|)^2}{4}\\ & = -\frac{(w^2 + y^2)^2}{2}\\ &\ge - \frac{(w^2 + y^2 + x^2 + z^2)^2}{2}\\ &= - \frac18 \end{align*} with equality if and only if $$x = z = 0$$ and $$y^2 + w^2 = 1/2$$ and $$2|yw| = w^2 + y^2 - 2|yw|$$ and $$yw \le 0$$ which results in $$x = z = 0, |yw| = 1/8$$. Then we have $$g = 1/64$$.

If $$yw > 0$$, we have \begin{align*} f &= 4yw (w^2 + y^2 + 2yw - x^2 - z^2 +2zx) \\ &\ge 4yw (w^2 + y^2 + 2yw - x^2 - z^2 - (x^2 + z^2))\\ &= 4yw (w^2 + y^2 + 2yw - (1 - 2y^2 - 2w^2))\\ &= 4yw (3w^2 + 3y^2 + 2yw - 1)\\ &\ge 4yw (6yw + 2yw - 1)\\ &= 32(yw - 1/16)^2 - 1/8\\ &\ge -1/8 \end{align*} with equality if and only if $$yw = 1/16$$ and $$y = w$$ and $$x + z = 0$$ and $$x^2 + y^2 + z^2 + w^2 = 1/2$$ which results in $$y^2 = w^2 = 1/16, x^2 = z^2 = 3/16$$. Then we have $$g = 1/64$$.

As a result, $$f_{\min} = -1/8$$ and at minimum $$abcd = 1/64$$.

My third solution.

We can assume that $$a > b > c > d$$.

Let $$P := (a-b)(b-c)(c-d)(d-a) = -(a-b)(b-c)(c-d)(a-d)$$.

We have \begin{align*} P &\ge - \frac{(a - b + c - d)^2}{4}(b - c)(a - d)\\ &= -\frac{(a - d - (b - c))^2}{4}(b - c)(a - d). \tag{1} \end{align*}

From $$a^2 + b^2 + c^2 + d^2 = 1$$, using $$(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$$, we have $$(a - d)^2 + (a + d)^2 + (b + c)^2 + (b - c)^2 = 2,$$ which results in $$a - d \le \sqrt{2 - (b - c)^2}. \tag{2}$$

Using (1) and (2), letting $$y := b - c$$, we have \begin{align*} P &\ge -\frac{(\sqrt{2 - y^2} - y)^2}{4}y \sqrt{2 - y^2} \\ &= \frac12 y^2(2 - y^2) - \frac12y\sqrt{2 - y^2}\\ &= \frac12(y\sqrt{2 - y^2} - 1/2)^2 - \frac18\\ &\ge -\frac18. \tag{3} \end{align*}

On the other hand, when $$a = \frac{\sqrt{3} + 1}{4}, b = \frac{\sqrt{3} - 1}{4}, c = -b, d = -a$$, we have $$P = -1/8$$. (Note: Solve the system of equations $$a + d = b + c = 0$$ and $$b - c = \frac{\sqrt{3} - 1}{2}$$ (from (3), $$y\sqrt{2 - y^2} - 1/2 = 0$$) and $$a^2 + b^2 + c^2 + d^2 = 1$$.)

Thus, the minimum is $$-1/8$$ with $$abcd = 1/64$$.

• Beautiful! (+1) Though I would like to note that the question was essentially APMO 22/5. Nice solutions can be found here.
– D S
Commented May 1 at 17:17
• @DS Thank for the information. I just searched APMO 2022 P5 solutions. Two solutions are given in apmo-official.org/static/solutions/apmo2022_sol.pdf Commented May 2 at 0:08