Let $a\geq b\geq c\geq d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3.$ If $a+b+c+d=6,$ prove that $d<0,36.$ If $a^2+b^2+c^2+d^2=14,$ prove that $(a+c)(b+d)\leq 8.$ When does equality hold?
Problem
Source: Romania JBMO TST 2022
Tags: algebra, romania, Romanian TST, inequalities
04.06.2022 03:44
oVlad wrote: Let $a\geq b\geq c\geq d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3.$ If $a+b+c+d=6,$ prove that $d<0,36.$ If $a^2+b^2+c^2+d^2=14,$ prove that $(a+c)(b+d)\leq 8.$ When does equality hold? $$d<0.36?$$
04.06.2022 04:22
sqing wrote: oVlad wrote: Let $a\geq b\geq c\geq d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3.$ If $a+b+c+d=6,$ prove that $d<0,36.$ If $a^2+b^2+c^2+d^2=14,$ prove that $(a+c)(b+d)\leq 8.$ When does equality hold? $$d<0.36?$$ Clearly.
04.06.2022 07:06
oVlad wrote: Let $a\geq b\geq c\geq d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3.$ If $a+b+c+d=6,$ prove that $d<0,36.$ If $a^2+b^2+c^2+d^2=14,$ prove that $(a+c)(b+d)\leq 8.$ When does equality hold? $a.$ $$ d \le \frac{3}{2}-\sqrt[4]{3}.$$ $b. \bullet 3=(a-b)(b-c)(c-d)(a-d) \le \left(\frac{a-b+b-c+c-d}{3} \right)^3(a-d) \rightarrow a-d \ge 3.$ $\bullet (a+c)(b+d)=a^2+b^2+c^2+d^2-\left[(a-b)^2+(b-c)^2+(c-d)^2+(a-b)(b-c)+(b-c)(c-d)+(c-d)(a-b) \right]$ $$\rightarrow (a+c)(b+d) \le a^2+b^2+c^2+d^2- \frac{2 \left(a-b+b-c+c-d \right)^2}{3}=a^2+b^2+c^2+d^2- \frac{2(a-d)^2}{3} \le 8.$$$$ \left(a=3;b=2;c=1;d=0 \rightarrow (a+c)(b+d)=8 \right).$$ .
04.06.2022 10:23
Intelligent
14.10.2023 15:39
any other solutions ??
14.10.2023 19:04
The official solution.
15.10.2023 04:09
Let $a, b, c, d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3 $ and $a^2+b^2+c^2+d^2=14 .$ Prove that $$(a+c)(b+d)\leq 8$$Equality holds when $a=-2,b=-1,c=0,d=-3 .$