AlgebraDifficulty 5.3AIME, harderProve itUnited States
Problem: Let a, b, c be positive real numbers. Assume that b19a19+c19b19+a19c19≤c19a19+a19b19+b19c19 Prove that b20a20+c20b20+a20c20≤c20a20+a20b20+b20c20
Solution
Solution: If we multiply the first equation by (abc)19 then it can be rewritten as (a19−b19)(b19−c19)(c19−a19)≤0. Similarly, the desired equation is equivalent to (a20−b20)(b20−c20)(c20−a20)≤0. These are evidently equivalent since a19−b19 and a20−b20 have the same sign, etc.
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