> Let ABC be a triangle P = (x:y:z) a point, A'B'C'

It's a cubic, 0=cyclic sum of

> the cevian triangle of a point Q = (u:v:w).

>

> Which is the locus of P such that the circles

> (APA'),(BPB'),(CPC') are coaxial?

(u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz>

It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

> Specialcase: P,Q are isogonal conjugate points.

>

--

> Antreas

Barry Wolk- 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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