## 19892Re: [EMHL] Re: Geometric inequality

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• Mar 5, 2011
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

> 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,
> > >
> > > 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
> > >
> >
>
>
>
>
> ------------------------------------
>