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1/a +3/b+5/c >= 4a^2 + 3b^2 + 2c^2 if a,b,c>0 with a^2 + b^2 + c^2 = 3

Source: 2018 Romania JBMO TST 1.2

June 19, 2020
algebrainequalities

Problem Statement

Let a,b,ca, b, c be positive real numbers such that a2+b2+c2=3a^2 + b^2 + c^2 = 3. Prove that 1a+3b+5c≥4a2+3b2+2c2\frac{1}{a}+\frac{3}{b}+\frac{5}{c} \ge 4a^2 + 3b^2 + 2c^2 When does the equality hold?
Marius Stanean
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