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Algebra Difficulty 4.4 AIME Prove it United States

Problem:
Let aa, bb, cc be positive real numbers with a+b+c=1a + b + c = 1. Prove that
a4+b4+c4abc. a^{4} + b^{4} + c^{4} \geq a b c .

Solution

Solution:
Since a+b+c=1a + b + c = 1, we can multiply the right side by a+b+ca + b + c to get the equivalent inequality
a4+b4+c4a2bc+ab2c+abc2. a^{4} + b^{4} + c^{4} \geq a^{2} b c + a b^{2} c + a b c^{2} .
By AM-GM,
2a4+b4+c44=a4+a4+b4+c44a4a4b4c44=a2bc. \frac{2 a^{4} + b^{4} + c^{4}}{4} = \frac{a^{4} + a^{4} + b^{4} + c^{4}}{4} \geq \sqrt[4]{a^{4} a^{4} b^{4} c^{4}} = a^{2} b c .
Similarly,
a4+2b4+c44ab2ca4+b4+2c44abc2 \begin{aligned} & \frac{a^{4} + 2 b^{4} + c^{4}}{4} \geq a b^{2} c \\ & \frac{a^{4} + b^{4} + 2 c^{4}}{4} \geq a b c^{2} \end{aligned}
Adding these three inequalities proves the desired inequality.

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