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19890Re: Geometric inequality

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  • RIJUL
    Mar 5, 2011
      Dear Franscisco,

      How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ??
      U,V,W and bc,ca,ab are not necessarily similarly sorted, aren't they?


      --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
      >
      > 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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