APMO 2022/5 wrote: 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. Mine . This problem was authored in (around) October 2019 after me attempting China MO 2019/1 and was motivated by the fact that I wanted an inequality where equality holds when all $a,b,c,d$ are distinct, and this version was finalized right before Indonesia RMM Training 2020. Overall, I'm actually quite surprised this appeared as a P5 because I initially proposed it with the intent that this would be in INAMO P8 or APMO P2. But apparently a lot of people seems to disagree. After the APMO was conducted, my friend told me that apparently the three variable version appeared before, and the method in that post is the official solution that I came up with. There are two solutions presented (the first and second official solution). The first one was mine, and the second one (a really slick approach) was communicated to me by Fajar Yuliawan. Solution 2, lower bound by Fajar Yuliawan Let us write $S(a,b,c,d) = (a - b)(b - c)(c - d)(d - a)$. First of all, the minimum value of $S(a,b,c,d)$ must be negative as $a < b < c < d \implies S(a,b,c,d) < 0$. Claim 01. The minimum value is achieved only if $a + b + c + d = 0$. Proof. Let us suppose otherwise, that a minimum value of $S(a,b,c,d)$ is achieved by a pair $(a,b,c,d)$ such that $a + b + c + d \not= 0$. Then write $\delta = \frac{a + b + c + d}{4}$, where $|\delta| > 0$. We have \begin{align*} \ell = \sum_{cyc} (a - \delta)^2 &= \sum_{cyc} a^2 - 2 \delta \sum_{cyc} a + 4 \delta^2 = \sum_{cyc} a^2 - 4\delta^2 < \sum_{cyc} a^2 \end{align*}Then by choosing the quadruple $(x,y,z,w) = \left( \frac{1}{\sqrt{\ell}} (a - \delta), \frac{1}{\sqrt{\ell}}(b - \delta), \frac{1}{\sqrt{\ell}}(c - \delta), \frac{1}{\sqrt{\ell}}(d - \ell) \right)$, we have \begin{align*} x^2 + y^2 + z^2 + w^2 &= \frac{1}{\ell} \cdot \sum_{cyc} (a - \delta)^2 = 1 \\ x + y + z + w &= \frac{1}{\sqrt{\ell}} \sum_{cyc} a - 4 \delta = 0 \\ S(x,y,z,w) &= \frac{1}{\ell} S(a,b,c,d) \stackrel{\ell < 1, S(a,b,c,d) < 0}{<} S(a,b,c,d) \end{align*}which contradicts the fact that $(a,b,c,d)$ achieves the minimum value of $S(a,b,c,d)$. Now, let us consider the substitution \begin{align*} x &= ac + bd \\ y &= ab + cd \\ z &= ad + bc \end{align*}Then, the original expression is just $(x - y)(x - z)$. Furthermore, we have $x,y,z \ge -\frac{1}{2}$ as \[ 2x + 1 = 2(ac + bd) + (a^2 + b^2 + c^2 + d^2) = (a + c)^2 + (b + d)^2 \ge 0 \]and the other inequalities analogously holds as well. Moreover, \[ 0 = (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ac + bd + ab + cd + ad + bc) = 1 + 2(x + y + z) \]Claim 02. Let $x,y,z \ge -\frac{1}{2}$ be real numbers such that $x + y + z = -\frac{1}{2}$. Then $(x - y)(x - z) \ge -\frac{1}{8}$, where equality holds if and only if $x = -\frac{1}{4}, \{ y,z \} = \left \{ \frac{1}{4}, -\frac{1}{2} \right \}$. Proof. We have \begin{align*} (x - y)(x - z) + \frac{1}{8} &= \left( 2y + z + \frac{1}{2} \right) \left( 2z + y + \frac{1}{2} \right) + \frac{1}{8} \\ &= \frac{1}{8} (4y + 4z + 1)^2 + \left( y + \frac{1}{2} \right) \left( z + \frac{1}{2} \right) \ge 0 \end{align*}Equality holds if and only if $y + z = -\frac{1}{4}$ and $-\frac{1}{2} \in \{ y,z \}$, which is what we wanted.

Too easy for an APMO P5 There are two cases to consider: If $a > b > c > d$, then \[ 8(a - b)(b - c)(c - d)(a - d) \le (a - b - c + d)^2 2(b - c)(a - d) \le \frac{1}{4} ((a - b - c + d)^2 + 2(b - c)(a - d))^2 \le \frac{1}{4} (2(a^2 + b^2 + c^2 + d^2))^2 = 1 \] If $b < c < d < a$, then do a similar thing as above.

Let $a-b=x_1$, $b-c=x_2$, $c-d=x_3$, we want to find the maximum of $x_1x_2x_3(x_1+x_2+x_3)$. Then there are two cases: $x_1x_3$ either bigger or equal than 0, or simply smaller than 0. Use calculus to obtain the final result.

The answer is $-\frac 18$, obtained when $(a,b,c,d)$ is a cyclic permutation of $( \frac{\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4})$ or $( \frac{\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4})$. Lemma. $a+b+c+d=0$. Proof 1. By translation. Note $(a,b,c,d)$ and $(a+k,b+k,c+k,d+k)$ produce the same value of $(a-b)(b-c)(c-d)(d-a)$. Let $k=a-d, l=b-d, m=c-d$ then I claim $d$ must be the vertex of $f(x)=x^2+(x+k)^2+(x+l)^2+(x+m)^2$. Assume $f(t)<f(d)=1$. Then $(\frac{t}{\sqrt{f(t)}},\frac{t+k}{\sqrt{f(t)}},\frac{t+l}{\sqrt{f(t)}}, \frac{t+m}{\sqrt{f(t)}})$ satisfy their squares being 1 and the desired expression is $(\frac{1}{f(t)})^2 (a-b)(b-c)(c-d)(d-a)$. Since $(a-b)(b-c)(c-d)(d-a)\le 0$, this value is better than that obtained in $a,b,c,d$. Therefore, $d$ is the vertex of the parabola. This means that $d=-\frac{k+l+m}{4}$, so $4d=d-a+d-b+d-c$, or $a+b+c+d=0$. Now we use the substitutions $x=a+b, -x=c+d, y=a-b, z=c-d$. This means we need to minimize $yz((\frac{y+z}{2})^2 - x^2)$. Furthermore, $2x^2+y^2+z^2=(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2=2(a^2+b^2+c^2+d^2)=2$ Substitute $x^2=\frac{2-y^2-z^2}{2}$, we need to minimize $\frac 14 yz((y+z)^2-4+2y^2+2z^2)$ Let $t=y+z, u=y-z$. Then $t^2+y^2=2(y^2+z^2)\le 4$. We need to minimize $\frac{1}{16} (t^2-u^2)(t^2-4+t^2+u^2)$ Let $p=t^2-u^2$. Case 1: $p>0$ then we minimize $t^2-4+t^2+u^2$, which happens when $u=0$. We need to find the vertex of the quadratic $\frac{1}{16}p(2p-4)$, which is $p=1$, when the value is $-\frac 18$. We have $t=\pm 1$ and $x=\pm \frac{\sqrt{3}}{2}$, yielding 4 solutions. Case 2: $p<0$ then we maximize $2t^2+u^2$. We should set $t^2+u^2=4$ so $2t^2+u^2-4=\frac 32 (t^2+u^2) + \frac 12 (t^2-u^2) - 4 = 2+\frac p2$. We have $t=\pm 1, u=\pm 3$ and $x=0$ yielding 4 solutions. The solutions are already described.

Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that$$ (a-b)(b-c)(c-d)(d-a)(ab+bc+cd+ad) \geq -1$$$$(a-b)(b-c)(c-d)(d-a) (ab+cd)\geq-\frac{1}{2}$$

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)(a+b+c+d) .$

This problem reminds me of IMO 2006 P3.

Gems98 wrote: This problem reminds me of IMO 2006 P3. No. This is actually more difficult

I love bash. We claim that $$\boxed{(a-b)(b-c)(c-d)(d-a) \geq -\frac{1}{8}}$$and equality is achieved at $$\boxed{(a,b,c,d)=\left( \frac{\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4} \right) \text{ or } \left( \frac{\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4} \right) \text{ or permutations}}$$ If any two adjacent numbers from $a,b,c,d,a$ are equal, then the expression is clearly $0$, so it satisfies the inequality. Thus assume $a \neq b \neq c \neq d \neq a$. Since the expression is cyclic, we can WLOG assume $a=\max \{a,b,c,d\}$. Thus $(a-b)(d-a)<0$, so the expression has the same sign as $-(b-c)(c-d)$. This is negative iff $c$ lies between $b$ and $d$, so we will assume that is the case. Also, switching $b,d$ doesn't change the expression, so WLOG we assume $b>d$. Thus we have $a>b>c>d$. Let $x=a-b$, $y=b-c$, and $z=c-d$, all positive numbers. The given expression becomes $-xyz(x+y+z)$, so it is sufficient to prove that $$xyz(x+y+z) \leq \frac18$$with the given condition $$d^2+(d+z)^2+(d+y+z)^2+(d+x+y+z)^2=1$$for some $d \in \mathbb{R}$. Let $$f(t)=t^2+(t+z)^2+(t+y+z)^2+(t+x+y+z)^2$$be a quadratic. This has its vertex at $t=-\frac{x+2y+3z}{4}$. Thus, by the above condition, its value at this vertex must be at most $1$. $$\implies \left(\frac{x+2y+3z}{4} \right)^2+\left(\frac{x+2y-z}{4} \right)^2+\left(\frac{-x+2y+z}{4} \right)^2+\left(\frac{3x+2y+z}{4} \right)^2 \leq 1$$Expanding this, we get $$3x^2+3z^2+4y^2+4xy+4xz+2xz \leq 4$$Let $m=\frac{x+z}{2}$. Note that $$3x^2+3z^2+2xz=2(x^2+z^2)+(x+z)^2 \geq 2(x+z)^2=8m^2$$Equality holds iff $x=z$. Putting this in the above inequality, we have $$y^2+2ym+2m^2 \leq 1$$Now the rest is just AM-GM: \begin{align*} xyz(x+y+z) & \leq y \cdot \left(\frac{x+z}{2} \right)^2(y+x+z) \\ &= ym^2(y+2m) \\ &= \frac12 \cdot 2m^2(y^2+2ym) \\ & \leq \frac 12 \cdot \frac{(y^2+2ym+2m^2)^2}{4} \\ \implies xyz(x+y+z) &\leq \frac18 \end{align*}as required. Equality holds when $x=z=m$, $2m^2=y^2+2ym=\frac12$, and also $d=-\frac{x+2y+3z}{4}$. These three conditions together give the first equality case, and switching $b,d$ around gives the second one. $\blacksquare$

Splendid, Supercali

Jalil_Huseynov wrote: 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 achived.

Supercali wrote: with the given condition $$d^2+(d+z)^2+(d+y+z)^2+(d+x+y+z)^2=1$$for some $d \in \mathbb{R}$. Let $$f(t)=t^2+(t+z)^2+(t+y+z)^2+(t+x+y+z)^2$$be a quadratic. This has its vertex at $t=-\frac{x+2y+3z}{4}$. Thus, by the above condition, its value at this vertex must be at most $1$. $$\implies \left(\frac{x+2y+3z}{4} \right)^2+\left(\frac{x+2y-z}{4} \right)^2+\left(\frac{-x+2y+z}{4} \right)^2+\left(\frac{3x+2y+z}{4} \right)^2 \leq 1$$$\blacksquare$ Can one explain me why at most 1, please ?

If $(a-b)(b-c)(c-d)(d-a)\ge 0$, then we get a lower bound of $0$. Hence assume $(a-b)(b-c)(c-d)(d-a)<0$. In this case, one of $(a-b)(c-d)$ or $(b-c)(d-a)$ must be negative so assume $(b-c)(d-a)<0$. Therefore $(b-c)(a-d)>0$ so \begin{align*} (a-b)(b-c)(c-d)(a-d)&\le\frac{1}{4}(b-c)(a-d)(a+c-b-d)^2\\ &=\frac{1}{8}\cdot 2(b-c)(a-d)(a+c-b-d)^2\\ &\le\frac{1}{32}[2(b-c)(a-d)+(a-d-b+c)^2]^2\\ &=\frac{1}{32}[(a-d)^2+(b-c)^2]^2\\ &=\frac{1}{32}[2(a^2+b^2+c^2+d^2)-(a+d)^2-(b+c)^2]^2\\ &\le\frac{1}{8}(a^2+b^2+c^2+d^2)^2. \end{align*}Therefore $(a-b)(b-c)(c-d)(d-a)\ge-\frac{1}{8}$. Now I will explain how to find equality cases. Equality holds, up to switching $(a,c)$, if and only if $a=-d$, $b=-c$ and $(a-d-b+c)^2=2(b-c)(a-d)$. Hence $(2a-2b)^2=2\cdot 2b\cdot 2a$, so $a^2-4ab+b^2=0$. A possible solution is $b=(2+\sqrt{3})a$. Hence $$1=a^2+b^2+c^2+d^2=2[1+(2+\sqrt{3})^2]a^2=(16+8\sqrt{3})a^2=4(1+\sqrt{3})^2a^2.$$Chossing $a=\frac{1}{2(1+\sqrt{3})}=\frac{\sqrt{3}-1}{4}$ works. In this case $b=\frac{\sqrt{3}+1}{4}$, $c=\frac{-\sqrt{3}-1}{4}$ and $d=\frac{-\sqrt{3}+1}{4}$. For ensurance, we can check that $a^2+b^2+c^2+d^2=1$ and $(a-b)(b-c)(c-d)(d-a)=-\frac{1}{8}$ (I will not do it). The other equality cases can be found easily.

Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that$$(a-b)(b-c)(c-d)(d-a) \geq -\frac{1}{8}$$$$\iff$$$$(a-b)(b-c)(c-d)\left((a-b)+(b-c)+(c-d)\right) \leq\frac{1}{8}$$

ali3985 wrote: Supercali wrote: with the given condition $$d^2+(d+z)^2+(d+y+z)^2+(d+x+y+z)^2=1$$for some $d \in \mathbb{R}$. Let $$f(t)=t^2+(t+z)^2+(t+y+z)^2+(t+x+y+z)^2$$be a quadratic. This has its vertex at $t=-\frac{x+2y+3z}{4}$. Thus, by the above condition, its value at this vertex must be at most $1$. $$\implies \left(\frac{x+2y+3z}{4} \right)^2+\left(\frac{x+2y-z}{4} \right)^2+\left(\frac{-x+2y+z}{4} \right)^2+\left(\frac{3x+2y+z}{4} \right)^2 \leq 1$$$\blacksquare$ Can one explain me why at most 1, please ? Because $1$ is in the image of $f$, it is at least the minimum.

We will show that the minimum possible value is $-\frac{1}{8}$. Without loss of generality assume that $a=\max \{a,b,c,d \}$. We may further assume that $(a-b)(b-c)(c-d)(d-a)<0$, since if this was not the case then we would be done. Since $d-a<0<a-b$, we have $(b-c)(c-d)>0$, and so either $b>c>d$ or $d>c>b$. Let $b>c>d$, with the other case being similar. Let $b-c=p$ and $a-d=q$, and so we have $$(a-b)(b-c)(c-d)(d-a)=-pq(a-b)(c-d) \geq -pq \frac{(a+c-b-d)^2}{4}=-\frac{pq(p-q)^2}{4}$$ Moreover, $$p^2+q^2=(b-c)^2+(a-d)^2=1-2(ad+bc) \leq 1+(a^2+d^2+b^2+c^2)=2,$$ and so $(p-q)^2=p^2+q^2 -2pq \leq 2-2pq$, which implies that $$-pq(p-q)^2 \geq -pq(2-2pq)=-2pq(1-pq) \geq -\frac{1}{2},$$ therefore $$(a-b)(b-c)(c-d)(d-a) \geq -\frac{pq(p-q)^2}{4} \geq -\frac{1}{8}$$ We now have to find all quadruples that give this minimum value. It is easy to obtain by examining when equality holds that the equality cases are $$(\frac{\sqrt{3}+1}{4},\frac{\sqrt{3}-1}{4},\frac{1-\sqrt{3}}{4},\frac{-1-\sqrt{3}}{4})$$ and $$(\frac{\sqrt{3}-1}{4},\frac{\sqrt{3}+1}{4},\frac{-1-\sqrt{3}}{4},\frac{1-\sqrt{3}}{4}),$$ along with all of their cyclic permutations.

WLOG $a$ is the largest. After analyzing signs, WLOG $a>b>c>d$ Suppose some $a > b > c > d$ achieves the minimum (this exists since the set of possible $(a,b,c,d)$ is compact). Define $w = \cfrac12(a-d), x = \cfrac12(b-c), y = \cfrac12(c-b), z = \cfrac12(d-a)$. Note that by the convexity of $x^2$, we have that $2w^2 < a^2 + d^2$. Summing this and other similar inequalities implies $$ w^2 + x^2 + y^2 + z^2 \le 1 $$However, by AM-GM also note that $$(w-x)^2, (y-z)^2 \ge |(a-b)(c-d)|$$and that $$(x-y)^2 = (b-c)^2, (w-z)^2=(a-d)^2$$ Multiplying the above, $$ |(w-x)(x-y)(y-z)(z-w)| \ge |(a-b)(b-c)(c-d)(d-a)| $$This would contradict our choice of $a, b, c, d$, unless we have equality and $w = a, x = b, y = c, z = d$ which further implies $a = -d, b = -c$ Now, $a^2 + b^2 = \frac12$ and we want to maximize $2a \cdot 2b \cdot (a-b)^2 = 2 \cdot 2ab \cdot (\frac12 - 2ab) \le \cfrac18$ Thus, the minimum value is $\cfrac{-1}8$ By analyzing the equality of our AM-GM, observe that $a = \cfrac{\sqrt{3}+1}{4}, b = \cfrac{\sqrt{3}-1}{4}$. Dealing with our WLOG statement, we find that the minimum values are obtained at permutations of $\left(\cfrac{\sqrt{3}+1}{4}, \cfrac{\sqrt{3}-1}{4}, \cfrac{-\sqrt{3}+1}{4}, \cfrac{-\sqrt{3}-1}{4} \right)$ or $\left( \cfrac{\sqrt{3}+1}{4}, \cfrac{-\sqrt{3}-1}{4}, \cfrac{-\sqrt{3}+1}{4}, \cfrac{\sqrt{3}-1}{4} \right)$

Jalil_Huseynov wrote: 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 achived. APMO 2022 Official Solution Solution

Jalil_Huseynov wrote: 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 achived. Who is the author?

mihaig wrote: Jalil_Huseynov wrote: 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 achived. Who is the author? Hi! See the first reply to this post (I don't know that user's actual name, but that's the author).

Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that$$ (a-1)(b-1)(c-1)(d-1)(a+b+c+d) \geq -\frac{81}{8}$$

GianDR wrote: mihaig wrote: Jalil_Huseynov wrote: 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 achived. Who is the author? Hi! See the first reply to this post (I don't know that user's actual name, but that's the author). Thank you. An example of professionalism

Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that$$ (a-bc)(b-cd)(c-da)(d-ab)(a+b+c+d) \geq-\frac{81}{128}$$

IndoMathXdZ wrote: APMO 2022/5 wrote: 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. Mine . This problem was authored in (around) October 2019 after me attempting China MO 2019/1 and was motivated by the fact that I wanted an inequality where equality holds when all $a,b,c,d$ are distinct, and this version was finalized right before Indonesia RMM Training 2020. Overall, I'm actually quite surprised this appeared as a P5 because I initially proposed it with the intent that this would be in INAMO P8 or APMO P2. But apparently a lot of people seems to disagree. After the APMO was conducted, my friend told me that apparently the three variable version appeared before, and the method in that post is the official solution that I came up with. [/hide] Bravo

sqing wrote: 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)(a+b+c+d) .$ Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that$$(a-b)(b-c)(c-d)(d-a)(a+b+c+d) > -\frac{3}{20}$$

Of course, any problem is correct when the proposer can provide a solution

Call $T=T(a,b,c,d)=(a-b)(b-c)(c-d)(d-a),$ notice that $T$ is unchanged by cyclic permutations of $a,b,c,d,$ so we can assume wlog that $a=\min (a,b,c,d).$ Moreover, we can see that $T(a,b,c,d)=T(a,d,c,b),$ so we can assume wlog that $b\le d.$ Let $m$ be the minimum value of $T$, since $T(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})=0,$ we get $m\le 0.$ Now if $c\ge d$ or $c\le b$ then $T\ge 0$ and so we are not interested in this case. We are left now with the case $a< b< c< d$ (any equality will result in $T=0,$ so we ignore these cases). Write $x=b-a,\;\;y=c-b$ and $z=d-c,$ then $x,y,z>0$ and the given identity becomes \[a^{2}+(a+x)^{2}+(a+x+y)^{2}+(a+x+y+z)^{2}=1\]and $T=-xyz(x+y+z).$ Now we apply Cauchy and AM-GM as follows: \[1=\Big( (-a)^{2}+(a+x+y+z)^{2}\Big) +\Big( (-a-x)^{2}+(a+x+y)^{2}\Big)\ge \frac{(x+y+z)^{2}+y^{2}}{2}\]\[=\frac{(x+y)^{2}+(y+z)^{2}}{2}+xz\ge (x+y)(y+z)+xz=y(x+y+z)+2xz\ge 2\sqrt{-2T}\]Hence, $T\ge -\frac{1}{8}$, and to achieve this, we need to fulfill the following system: \[\begin{cases} a+d=b+c=0\;\;(\Rightarrow x=z,\;\;2x+y=-2a) \\ a^{2}+(a+x)^{2}=\frac{1}{2} \\ y(2x+y)=2x^{2} \\ \end{cases} \]we will get $(a,b,c,d)=(-\frac{\sqrt{3}+1}{4},-\frac{\sqrt{3}-1}{4},\frac{\sqrt{3}-1}{4},\frac{\sqrt{3}+1}{4})=p.$ So the equality cases are only the cyclic permutations of $p$ and $p'=(d,c,b,a)$.

........

mihaig wrote: Gems98 wrote: This problem reminds me of IMO 2006 P3. No. This is actually more difficult. If one is not an examinee, this problem, in effect, is not so difficult. (I mean, one is capable of availing oneself of diverse measures to the best of one's ability.) For instance: \[\begin{aligned}(a-b)(b-c)(c-d)(d-a)&=\frac{{\left(ab+bc+cd+da-ac-bd\right)}^2+{\left(ac-bd\right)}^2}2+\frac{{\left(a^2+c^2-b^2-d^2\right)}^2-{\left(a^2+b^2+c^2+d^2\right)}^2}8\textnormal,\\&\geqq-\frac18{\left(a^2+b^2+c^2+d^2\right)^2}\textnormal{…}\end{aligned}\]Done! (by 陳晨 )

I'm actually curious how this got selected as a problem 5 because even the author intended this as a problem 2 lol.

Rhapsodies_pro wrote: mihaig wrote: Gems98 wrote: This problem reminds me of IMO 2006 P3. No. This is actually more difficult If one is not an examinee, this problem, in effect, is not so difficult. (I mean, one is capable of availing oneself of diverse measures to the best of one's ability.) For instance: \[\begin{aligned}(a-b)(b-c)(c-d)(d-a)&=\frac{{\left(ab+bc+cd+da-ac-bd\right)}^2+{\left(ac+bd\right)}^2}2+\frac{{\left(a^2-b^2+c^2-d^2\right)}^2-{\left(a^2+b^2+c^2+d^2\right)}^2}8\textnormal,\\&\geqq-\frac18{\left(a^2+b^2+c^2+d^2\right)^2}\textnormal{…}\end{aligned}\]Done! (by 陳晨 ) Oh my lord, what a wonderful solution! How did you motivate this! If it is something, the author of this solution is not human.

Indeed, it is

Solution. We only deal with the case $a>b>c>d$. Denote $T = (a-b)(b-c)(c-d)(d-a)$. Note that $T$ won't change if we change the pers of $a,b,c,d$ (this is used for the last part: equality cases). Also, if we denote $x = a-b, y = b-c, z = c-d \implies d-a = -(x+y+z)$. So basically we want the minimum of $-xyz(x+y+z)$, given that $(x+y+z+d)^2+(y+z+d)^2+(z+d)^2+d^2=1$. Here, negative terms are especially hard to tackle, so we consider how to create negative ones on our own or to cancel out them. After some racking I realized we can manipulate on the squares: $d^2 = (-d)^2, (x+d)^2 = (-x-d)^2$. Wonderful things immediately turn out then: we may apply C-S or something like that. \begin{align*} 1&=\left((-d)^{2}+(x+y+z+d)^{2}\right) +\left( (-x-d)^{2}+(x+y+d)^{2}\right) \\ & \geqslant \frac{(x+y+z)^{2}+y^{2}}{2} \\ & =\frac{(x+y)^2+(y+z)^2+2xz}{2} \\ & \geqslant (x+y)(y+z)+xz \qquad \text{(AM-GM)}\\ & = y(x+y+z)+2xz \geqslant 2\sqrt{-2T} \\ & \implies T\leqslant \frac 18. \end{align*}Equality case: permutations of $\left(\cfrac{\sqrt{3}+1}{4}, \cfrac{\sqrt{3}-1}{4}, \cfrac{-\sqrt{3}+1}{4}, \cfrac{-\sqrt{3}-1}{4} \right)$ or $\left( \cfrac{\sqrt{3}+1}{4}, \cfrac{-\sqrt{3}-1}{4}, \cfrac{-\sqrt{3}+1}{4}, \cfrac{\sqrt{3}-1}{4} \right)$. Remarks: Well my solution is almost the same as Zezohabibullah's. (Why does my sol always bumps into others'??) PS Mr. Chen Chen's beautiful solution really amazed me! This problem is not hard. Once you're familiar with C-S and AM-GM common manipulations it's easy to come up with the above solution. BTW don't try to figure out the equality case as soon as you meet this problem. Analyze it first. (just think of why it's placed in #5 lol)

Beautiful

One of the nice solutions (Also done by CANBANKAN) : The answer is $-\frac 18$, obtained when $(a,b,c,d)$ is a cyclic permutation of $( \frac{\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4})$ or $( \frac{\sqrt{3}+1}{4}, \frac{-\sqrt{3}-1}{4}, \frac{-\sqrt{3}+1}{4}, \frac{\sqrt{3}-1}{4})$. Lemma. $a+b+c+d=0$. Proof 1. By translation. Note $(a,b,c,d)$ and $(a+k,b+k,c+k,d+k)$ produce the same value of $(a-b)(b-c)(c-d)(d-a)$. Let $k=a-d, l=b-d, m=c-d$ then I claim $d$ must be the vertex of $f(x)=x^2+(x+k)^2+(x+l)^2+(x+m)^2$. Assume $f(t)<f(d)=1$. Then $(\frac{t}{\sqrt{f(t)}},\frac{t+k}{\sqrt{f(t)}},\frac{t+l}{\sqrt{f(t)}}, \frac{t+m}{\sqrt{f(t)}})$ satisfy their squares being 1 and the desired expression is $(\frac{1}{f(t)})^2 (a-b)(b-c)(c-d)(d-a)$. Since $(a-b)(b-c)(c-d)(d-a)\le 0$, this value is better than that obtained in $a,b,c,d$. Therefore, $d$ is the vertex of the parabola. This means that $d=-\frac{k+l+m}{4}$, so $4d=d-a+d-b+d-c$, or $a+b+c+d=0$. Now we use the substitutions $x=a+b, -x=c+d, y=a-b, z=c-d$. This means we need to minimize $yz((\frac{y+z}{2})^2 - x^2)$. Furthermore, $2x^2+y^2+z^2=(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2=2(a^2+b^2+c^2+d^2)=2$ Substitute $x^2=\frac{2-y^2-z^2}{2}$, we need to minimize $\frac 14 yz((y+z)^2-4+2y^2+2z^2)$ Let $t=y+z, u=y-z$. Then $t^2+y^2=2(y^2+z^2)\le 4$. We need to minimize $\frac{1}{16} (t^2-u^2)(t^2-4+t^2+u^2)$ Let $p=t^2-u^2$. Case 1: $p>0$ then we minimize $t^2-4+t^2+u^2$, which happens when $u=0$. We need to find the vertex of the quadratic $\frac{1}{16}p(2p-4)$, which is $p=1$, when the value is $-\frac 18$. We have $t=\pm 1$ and $x=\pm \frac{\sqrt{3}}{2}$, yielding 4 solutions. Case 2: $p<0$ then we maximize $2t^2+u^2$. We should set $t^2+u^2=4$ so $2t^2+u^2-4=\frac 32 (t^2+u^2) + \frac 12 (t^2-u^2) - 4 = 2+\frac p2$. We have $t=\pm 1, u=\pm 3$ and $x=0$ giving 4 solutions. I really like this solution (Previously done by CANBANKAN). I have also done APMO Problem 5 by same technique and got this while checking on AoPS.

Nice problem without any advanced methods. i heckin loooooooove smoothing!!! The answer is $-\frac{1}{8}$, acheived by $\left(\frac{\sqrt{3}+1}{4},\frac{\sqrt{3}-1}{4},\frac{1-\sqrt{3}}{4},\frac{-1-\sqrt{3}}{4}\right)$, its reverse, and cyclic permutations. First, note that the minimum value is negative, since we can obtain a negative number by choosing $a\geq b\geq c\geq d$. Actually, note that cyclic permutations and reversals of $(a,b,c,d)$ won't change $(a-b)(b-c)(c-d)(d-a)$, and the two other equivalence classes $a\geq b\geq d\geq c$ and $a\geq c\geq b\geq d$ both result in a positive value of $(a-b)(b-c)(c-d)(d-a)$, so we may WLOG assume that $a\geq b\geq c\geq d$. Now we relax the condition to $a^2+b^2+c^2+d^2 \leq 1$. Note that this doesn't actually change anything regarding the minimum, since if we do find a minimal (negative) tuple $(a,b,c,d)$ with $a^2+b^2+c^2+d^2=k<1$, then we can multiply each term by $1/\sqrt{k}>1$, which decreases the value of $(a-b)(b-c)(c-d)(d-a)$. Observe that shifting every variable by the same constant $t$ will not change the value of $(a-b)(b-c)(c-d)(d-a)$. Thus if $(a,b,c,d)$ is minimal, we must have $(a+t)^2+(b+t)^2+(c+t)^2+(d+t)^2\geq 1$ for all $t$, by our previous result. Since we also have $a^2+b^2+c^2+d^2=1$, this means that $$4t^2+2t(a+b+c+d)\geq 0.$$Viewing this as a quadratic in $t$, for this to be always true we must have $a+b+c+d=0$. Keeping in mind $a\geq b \geq c \geq d$, we employ the key substitution \begin{align*} a&=k+x\\ b&=k-x\\ c&=-k+y\\ d&=-k-y \end{align*}where $x,y \geq 0$ and $b\geq c \iff 2k\geq x+y$. We then want to minimize $$(2x)(-(x+y)+2k)(2y)(-(x+y)-2k)=-4xy(4k^2-(x+y)^2),$$subject to $4k^2+2x^2+2y^2=1$. Substituting $4k^2=1-2x^2-2y^2$, we simply want to maximize $$4xy(1-3x^2-3y^2-2xy).$$Note that there are a number of conditions from our substitution tied to this expression, but all we need is $x,y\geq 0$. Then, we substitute $p=x^2+y^2$ and $q=xy\geq 0$, this time using the condition $p\geq 2q$ by AM-GM, so the expression becomes $$4q(1-3p-2q).$$Since $4q\geq 0$, we should minimize $3p$, so the maximum occurs when $p=2q$ in which case the expression is equivalent to $4q(1-8q)$. This is a quadratic in $q$ whose maximum of $\frac{1}{8}$ (corresponding to the original minimum of $-\frac{1}{8}$) occurs when $q=\frac{1}{16}$, so then $p=\frac{1}{8}$. From this we then extract $x=y=\frac{1}{4}$, so $k=\frac{\sqrt{3}}{4}$ and we obtain the described solutions, as desired. $\blacksquare$ Remark: The same key substitution appears in my solution to ISL 2005/A3, where given $p+q+r+s=9$ and $p\geq q\geq r\geq s$ we substitute \begin{align*} p&=\frac{9}{4}+k+x\\ q&=\frac{9}{4}+k-x\\ r&=\frac{9}{4}-k+y\\ s&=\frac{9}{4}-k-y. \end{align*}This substitution seems to be quite powerful, especially on these "non-traditional" inequalities that are more smoothing-flavored and less symmetrical to begin with, so I find it rather strange that I haven't really seen it.

........

A bit easy.

Solved with CT17. Assume WLOG that $a \ge b,c,d$ and $b \ge d$. If $(a-b)(b-c)(c-d)(d-a)$ is positive, then we must have $a>b>c>d$, so assume WLOG that this is true. Let $x=(a-b)$, $y=(b-c)$, and $z=c-d$. Notice that \begin{align*} 4 &= 4a^2+4b^2+4c^2+4d^2 \\ &= (a+b+c+d)^2+(a-b)^2+(b-c)^2+(c-d)^2+(a-c)^2+(b-d)^2+(a-d)^2 \\ &\ge x^2+y^2+z^2+(x+y)^2+(y+z)^2+(x+y+z)^2 \\ &= 3x^2+3z^2+2xz+4y^2+4yx+4yz \\ &= 2(x^2+z^2)+(x+z)^2+4y^2+4y(x+z) \end{align*}and we want to maximize $xyz(x+y+z)$. If $2(x^2+z^2)+(x+z)^2+4y^2+4y(x+z)<4$, we can scale $x$, $y$, and $z$ up to increase the value of $xyz(x+y+z)$, so assume WLOG that $2(x^2+z^2)+(x+z)^2+4y^2+4y(x+z)=4$. If $x \ne z$, we can move $x$ and $z$ closer while keeping $x+z$ constant and increase $y$ accordingly to keep the condition satisfied, which increases $xyz(x+y+z)$. Thus, we can assume that $x=z$, so we want to maximize $2x^3y+x^2y^2$ under $2x^2+2xy+y^2=1$. Let $k=\tfrac{y}{x}$, so we want to maximize $x^4(k^2+2k)$ under $x^2(k^2+2k+2)=1$. This is equivalent to maximizing $\tfrac{k^2+2k}{(k^2+2k+2)^2}$ for positive real numbers $k$. Notice that the range of $m=k^2+2k$ for positive real $k$ is the positive real numbers, so we want to maximize $\tfrac{m}{(m+2)^2}$. We have $\tfrac{(m+2)^2}{4}=\tfrac{m^2+2m+2m+4}{4} \ge 2m$ by AM-GM, so $\tfrac{m}{(m+2)^2} \le \tfrac{1}{8}$, with equality when $m=2$ and thus $k=\sqrt{3}-1$. This gives $x=z=\tfrac{1}{2}$ and $y=\tfrac{\sqrt{3}-1}{2}$. Since equality holds only when $a+b+c+d=0$, we have $(a,b,c,d)=(\tfrac{\sqrt{3}+1}{4},\tfrac{\sqrt{3}-1}{4},\tfrac{-\sqrt{3}+1}{4},\tfrac{-\sqrt{3}-1}{4})$. Lifting the WLOG assumptions, the minimum value $-\tfrac{1}{8}$ of $(a-b)(b-c)(c-d)(d-a)$ is achieved when either $(a,b,c,d)$ or $(d,c,b,a)$ is a cyclic permutation of $(\tfrac{\sqrt{3}+1}{4},\tfrac{\sqrt{3}-1}{4},\tfrac{-\sqrt{3}+1}{4},\tfrac{-\sqrt{3}-1}{4})$. $\square$