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Pythagorean means

Authors
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    Name
    Vu Hung
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Problem Statement

For 0<a<b0<a<b, define

A=a+b2,G=ab,H=2aba+b.A=\frac{a+b}{2},\quad G=\sqrt{ab},\quad H=\frac{2ab}{a+b}.
  • Prove H<G<AH<G<A.
  • Find ba\frac ba such that a,G,Aa,G,A is an AP.
  • Show H,G,AH,G,A cannot be an AP unless a=ba=b.

Hints

Use (a−b)2>0(\sqrt a-\sqrt b)^2>0 and G2=AHG^2=AH.

Solutions

From (a−b)2>0(\sqrt a-\sqrt b)^2>0, get A>GA>G. Also

AH=a+b2⋅2aba+b=ab=G2,AH=\frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab=G^2,

so H,G,AH,G,A form a GP and therefore H<G<AH<G<A. For a,G,Aa,G,A in AP:

2G=a+A  ⟹  4ab=3a+b.2G=a+A \implies 4\sqrt{ab}=3a+b.

Let x=b/ax=b/a; then (x−9)(x−1)=0(x-9)(x-1)=0, and x>1x>1 gives x=9x=9. If H,G,AH,G,A were AP, then 2G=H+A=G2/A+A2G=H+A=G^2/A+A, forcing A=GA=G, contradiction for a≠ba\ne b.

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