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Algebra Difficulty 6.1 National Olympiad Prove it Saudi Arabia

Let x1x_1, x2x_2, x3x_3 and y1y_1, y2y_2, y3y_3 are 6 positive real numbers such that
x1+x2+x3=y1y2y3 and y1+y2+y3=x1x2x3. x_1 + x_2 + x_3 = y_1y_2y_3 \text{ and } y_1 + y_2 + y_3 = x_1x_2x_3.
Find the minimum value of T=x1y1+x2y2+x3y3T = x_1y_1 + x_2y_2 + x_3y_3.

Solution

Using AM-GM, we have
x1+x2+x3≥3x1x2x33  ⟹  y1y2y3≥3x1x2x33. x_1 + x_2 + x_3 \ge 3\sqrt[3]{x_1x_2x_3} \implies y_1y_2y_3 \ge 3\sqrt[3]{x_1x_2x_3}.
Similarly, x1x2x3≥3y1y2y33x_1x_2x_3 \ge 3\sqrt[3]{y_1y_2y_3}. Multiplying these inequalities, side-by-side, we get
(x1x2x3⋅y1y2y3)23≥9  ⟹  x1x2x3⋅y1y2y3≥27. \sqrt[3]{(x_1x_2x_3 \cdot y_1y_2y_3)^2} \ge 9 \implies x_1x_2x_3 \cdot y_1y_2y_3 \ge 27.
Now applying AM-GM for the given expression
T≥3x1x2x3⋅y1y2y33≥3273=9. T \ge 3\sqrt[3]{x_1x_2x_3 \cdot y_1y_2y_3} \ge 3\sqrt[3]{27} = 9.
Hence, the minimum value is 9, equality occurs when
x1=x2=x3=y1=y2=y3=3. x_1 = x_2 = x_3 = y_1 = y_2 = y_3 = \sqrt{3}.

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