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Algebra Difficulty 4.6 AIME Prove it JBMO

Problem:
Let a,b,c,d,ea, b, c, d, e be real numbers such that a+b+c+d+e=0a+b+c+d+e=0. Let, also A=ab+bc+cd+de+eaA=ab+bc+cd+de+ea and B=ac+ce+eb+bd+daB=ac+ce+eb+bd+da.
Show that
2005A+B≤0 or A+2005B≤0 2005 A+B \leq 0 \text{ or } \quad A+2005 B \leq 0

Solution

Solution:
We have
0=(a+b+c+d+e)2=a2+b2+c2+d2+e2+2A+2B 0=(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2+2A+2B
This implies that
A+B≤0 or 2006(A+B)=(2005A+B)+(A+2005B)≤0 A+B \leq 0 \text{ or } 2006(A+B)=(2005 A+B)+(A+2005 B) \leq 0
This implies the conclusion.

We have
2A+2B=a(b+c+d+e)+b(c+d+e+a)+c(d+e+a+b)+d(e+a+b+c)+e(a+b+c+d)=−a2−b2−c2−d2−e2≤0 \begin{aligned} 2A+2B &= a(b+c+d+e)+b(c+d+e+a)+c(d+e+a+b) \\ &\quad +d(e+a+b+c)+e(a+b+c+d) \\ &= -a^2-b^2-c^2-d^2-e^2 \leq 0 \end{aligned}
Therefore we have A+B≤0A+B \leq 0, etc.

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