Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it United States

Problem:
Let aa, bb, cc be positive real numbers. Assume that
a19b19+b19c19+c19a19a19c19+b19a19+c19b19 \frac{a^{19}}{b^{19}} + \frac{b^{19}}{c^{19}} + \frac{c^{19}}{a^{19}} \leq \frac{a^{19}}{c^{19}} + \frac{b^{19}}{a^{19}} + \frac{c^{19}}{b^{19}}
Prove that
a20b20+b20c20+c20a20a20c20+b20a20+c20b20 \frac{a^{20}}{b^{20}} + \frac{b^{20}}{c^{20}} + \frac{c^{20}}{a^{20}} \leq \frac{a^{20}}{c^{20}} + \frac{b^{20}}{a^{20}} + \frac{c^{20}}{b^{20}}

Solution

Solution:
If we multiply the first equation by (abc)19(a b c)^{19} then it can be rewritten as
(a19b19)(b19c19)(c19a19)0. \left(a^{19} - b^{19}\right)\left(b^{19} - c^{19}\right)\left(c^{19} - a^{19}\right) \leq 0.
Similarly, the desired equation is equivalent to
(a20b20)(b20c20)(c20a20)0. \left(a^{20} - b^{20}\right)\left(b^{20} - c^{20}\right)\left(c^{20} - a^{20}\right) \leq 0.
These are evidently equivalent since a19b19a^{19} - b^{19} and a20b20a^{20} - b^{20} have the same sign, etc.

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