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Algebra Difficulty 4.9 AIME Prove it Austria

Determine the largest constant CC such that
(x1+x2++x6)2C(x1(x2+x3)+x2(x3+x4)++x6(x1+x2)) (x_1 + x_2 + \dots + x_6)^2 \geq C \cdot (x_1(x_2 + x_3) + x_2(x_3 + x_4) + \dots + x_6(x_1 + x_2))
holds for all real numbers x1,x2,,x6x_1, x_2, \dots, x_6.
For this CC, determine all x1,x2,,x6x_1, x_2, \dots, x_6 such that equality holds.

Solution

We rewrite the right-hand side

Expanding yields
X2+Y2+Z2XY+YZ+ZX X^2 + Y^2 + Z^2 \geq XY + YZ + ZX
This is equivalent to
(XY)2+(YZ)2+(ZX)20 (X - Y)^2 + (Y - Z)^2 + (Z - X)^2 \geq 0
with equality for XY=YZ=ZX=0X - Y = Y - Z = Z - X = 0, i.e., X=Y=ZX = Y = Z, thus x1+x4=x2+x5=x3+x6x_1 + x_4 = x_2 + x_5 = x_3 + x_6.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.