Problem: Let a, b, c be positive real numbers with a+b+c=1. Prove that a4+b4+c4≥abc.
Solution
Solution: Since a+b+c=1, we can multiply the right side by a+b+c to get the equivalent inequality a4+b4+c4≥a2bc+ab2c+abc2. By AM-GM, 42a4+b4+c4=4a4+a4+b4+c4≥4a4a4b4c4=a2bc. Similarly, 4a4+2b4+c4≥ab2c4a4+b4+2c4≥abc2 Adding these three inequalities proves the desired inequality.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.