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a,b,c in HP; (a^2+b^2) etc in GP [RMO2-2011, India]

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December 31, 2011
inequalitiesarithmetic sequencegeometric sequence

Problem Statement

Let a,b,c>0.a,b,c>0. If 1a,1b,1c\frac 1a,\frac 1b,\frac 1c are in arithmetic progression, and if a2+b2,b2+c2,c2+a2a^2+b^2,b^2+c^2,c^2+a^2 are in geometric progression, show that a=b=c.a=b=c.
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