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

> > >

> >

>