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## 19891Re: Geometric inequality

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• Mar 5, 2011
Rijul, you are right, I didn't take this into account.

My apologies.

--- In Hyacinthos@yahoogroups.com, "RIJUL" <rijul.champ@...> wrote:
>
> 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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