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? $$d<0.36?$$

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.

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? $a.$ $$ d \le \frac{3}{2}-\sqrt[4]{3}.$$ $b. \bullet 3=(a-b)(b-c)(c-d)(a-d) \le \left(\frac{a-b+b-c+c-d}{3} \right)^3(a-d) \rightarrow a-d \ge 3.$ $\bullet (a+c)(b+d)=a^2+b^2+c^2+d^2-\left[(a-b)^2+(b-c)^2+(c-d)^2+(a-b)(b-c)+(b-c)(c-d)+(c-d)(a-b) \right]$ $$\rightarrow (a+c)(b+d) \le a^2+b^2+c^2+d^2- \frac{2 \left(a-b+b-c+c-d \right)^2}{3}=a^2+b^2+c^2+d^2- \frac{2(a-d)^2}{3} \le 8.$$$$ \left(a=3;b=2;c=1;d=0 \rightarrow (a+c)(b+d)=8 \right).$$ .

Intelligent

any other solutions ??

The official solution.

Good.

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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 $a=-2,b=-1,c=0,d=-3 .$