National Competition

Algebra Difficulty 4.9 AIME Prove it Austria

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

Solution

We rewrite the right-hand side

Expanding yields
$X^2 + Y^2 + Z^2 \geq XY + YZ + ZX$
This is equivalent to
$(X - Y)^2 + (Y - Z)^2 + (Z - X)^2 \geq 0$
with equality for $X - Y = Y - Z = Z - X = 0$ , i.e., $X = Y = Z$ , thus $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.