## 19890Re: Geometric inequality

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• 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.
>
>
>
> >
> > 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