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19889Re: Geometric inequality

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  • Francisco Javier
    Mar 5, 2011
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      Dear Nikos,

      By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have

      U(bc)^2 + V(ca)^2 + W(ab)^2
      >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
      >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),

      hence is enough to prove that

      (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,

      but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to

      (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),

      which is always positive.



      --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
      >
      > Dear friends,
      > does anybody can give a simple proof to the
      > following inequality?
      > If p, q, r are arbitrary positive real numbers
      > a,b,c the sides, F the area of triangle ABC
      > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
      > then prove that
      > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
      >
      > Best regards
      > Nikos Dergiades
      >
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