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19892Re: [EMHL] Re: Geometric inequality

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  • Nikolaos Dergiades
    Mar 5, 2011
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      Dear Franscisco,
      I knew this approach but it is not
      always correct because
      Chebyshev's Inequality
      3(a1b1 + a2b2 + a3b3) >= (a1 + a2 + a3)(b1 + b2 + b3)
      holds as said RIJUL only when
      a1 > a2 > a3 we have b1 > b2 > b3.

      Best regards
      Nikos Dergiades

      > 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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