Let $a,b,c,d$ be real numbers with $0\leqslant a,b,c,d\leqslant 1$. Prove that $$ab(a-b)+bc(b-c)+cd(c-d)+da(d-a)\leqslant \frac{8}{27}.$$ Find all quadruples $(a,b,c,d)$ of real numbers with $0\leqslant a,b,c,d\leqslant 1$ for which equality holds in the above inequality.
Problem
Source: Benelux MO 2019 P1
Tags: inequalities, 2019
WolfusA
28.04.2019 17:36
A Given expression equals to $(a-c)(b-d)(a+c-b-d)$ and is cyclic so assume $a\ge b,c,d$. By AM, GM $(b-d)(a+c-b-d)\le \frac{(a+c-2d)^2}{4}$ and $(a-c)\frac{(a+c-2d)^2}{4}\le \frac{8(a-d)^3}{27}\le \frac{8}{27}$ B With given assumption that $a$ is the biggest of variables we have equality iff $a-d=1, b-d=a+c-b-d, a-c=\frac{a+c-2d}{2}$ which is $a=1,d=0,b=\frac23, c=\frac13$ Of course other cyclic permutations is also the equality condition when we are free from assuming which variable is the biggest.
WolfusA
28.04.2019 18:54
Corrected. My mistake was writing down $a-c=1$ instead of $a-d=1$
sqing
29.04.2019 03:29
Let $a_1,a_2,\cdots,a_n \in [0,1] (n\ge 3).$ Prove or disprove $$a_1a_2(a_1-a_2)+a_2a_3(a_2-a_3)+\cdots+a_{n-1}a_n(a_{n-1}-a_n)+a_na_1(a_n-a_1)\leq\left(\frac{n-2}{n-1}\right)^{n-1}.$$
Lepuslapis
29.04.2019 11:49
sqing wrote: Let $a_1,a_2,\dots,a_n \in [0,1] (n\geqslant 3).$ Prove or disprove $$a_1a_2(a_1-a_2)+a_2a_3(a_2-a_3)+\cdots+a_{n-1}a_n(a_{n-1}-a_n)+a_na_1(a_n-a_1)\leqslant\left(\frac{n-2}{n-1}\right)^{n-1}.$$ Unfortunately, this does not hold. E.g., for $n=6$, letting $a=0,b=1,c=1/2,d=0,e=1,f=1/2$ gives $ab(a-b)+bc(b-c)+cd(c-d)+de(d-e)+ef(e-f)+fa(f-a)=1/2>(4/5)^5$.
sqing
29.04.2019 12:08
Lepuslapis wrote:
sqing wrote:
Let $a_1,a_2,\dots,a_n \in [0,1] (n\geqslant 3).$ Prove or disprove $$a_1a_2(a_1-a_2)+a_2a_3(a_2-a_3)+\cdots+a_{n-1}a_n(a_{n-1}-a_n)+a_na_1(a_n-a_1)\leqslant\left(\frac{n-2}{n-1}\right)^{n-1}.$$
Unfortunately, this does not hold. E.g., for $n=6$, letting $a=0,b=1,c=1/2,d=0,e=1,f=1/2$ gives $ab(a-b)+bc(b-c)+cd(c-d)+de(d-e)+ef(e-f)+fa(f-a)=1/2>(4/5)^5$.
NaPrai wrote:
Answer. The maximum value of $\sum_{cyc}ab(a-b)$ is $\frac{1}{4}$, which can be achieved by take $(a,b,c)=\left(1,\frac{1}{2},0\right)$ or its cyclic permutation.
Solution. Note that \begin{align*}\sum_{cyc}ab(a-b)=(a-b)(b-c)(a-c).\end{align*}So, we will divide the solution into two cases as following.
Case I: $a \le b \le c$ or $b \le c \le a$ or $c \le a \le b$. For this case, it's trivial that $\sum_{cyc}ab(a-b) \le 0 < \frac{1}{4}$.
case II: $a \ge b \ge c$ or $b \ge c \ge a$ or $c \ge a \ge b$. Due to the cyclic, we may assume WLOG that $a \ge b\ge c$.
By AM-GM, we have \begin{align*}(a-b)(b-c)(a-c) \le \frac{1}{4}(a-c)^3 \le \frac{1}{4}.\end{align*}
strong_boy
05.11.2022 22:53
I love this problem It is easy to see $\sum ab(a-b) = (b-d)(a-c)(a+c-d-b)$ . Now Let $A= (b-d)(a-c)(a+c-d-b)$ . Case 1 : If $b+d \geq 0$ : Case 1.1 : If $a+c-b-d < 0$ then $A \leq 0$ . Case 1.2 : If $a+c-b-d \geq 0 $ then : $\sqrt[3]{A} = \sqrt[3]{(b-d)(a-c)(a+c-d-b)} \leq \frac{2d-2c}{3}$ . And it is easy to see $\frac{2d-2c}{3} \leq \frac{3}{2}$ . ($d-c \leq 1$) So ! . Let's see second case : Case 2 : If $b+d < 0$ : Case 1.1 : If $a+c-b-d \geq 0$ then $A \leq 0$ . Case 1.2 : If $a+c-b-d < 0 $ then : $\sqrt[3]{A} = \sqrt[3]{(b-d)(a-c)(a+c-d-b)} \leq \frac{2d-2c}{3}$ . And it is easy to see $\frac{2d-2c}{3} \leq \frac{3}{2}$ . ($d-c \leq 1$)