MathDB
maximum of

Source: 2016 KJMO #7

November 13, 2016
algebrainequalities

Problem Statement

positive integers a1,a2,...,a9a_1, a_2, . . . , a_9 satisfying a1+a2+...+a9=90a_1+a_2+ . . . +a_9 =90 find maximum of 1a12a2...9a9a1!a2!...a9!\frac{1^{a_1} \cdot 2^{a_2} \cdot . . . \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot . . . \cdot a_9!}
I was really shocked because there are no inequality problems at KJMO and the test difficulty even more lower...