Re: [obm-l] algebrismos

Sauda, c~oes,

Segue uma solução.

[]'s

Luis

>From: "ndergiades"

>Reply-To: Hyacinthos@yahoogroups.com
>To: Hyacinthos@yahoogroups.com
>Subject: Re: [EMHL] S and more
>Date: Sat, 18 May 2002 23:53:38 -0000
>
>Dear Luvs,
>
>You wrote:
>
> >And this one, taken from another list:
> >
> >Prove que (Show that, standard notation)
> >
> >(a^2 + b^2)/2m_c + (b^2 + c^2)/2m_a + (a^2 + c^2)/2m_b <=
>6R
>
>******
>If m1, m2, m3 are the medians of ABC then your inequality
>is written
>(bb+cc)/2m1 + (cc+aa)/2m2 + (aa+bb)/2m3 <= 6R.
>
>From 9RR - (aa + bb + cc) = OH^2 we get
>aa + bb + cc <= 9RR
>Since (x + y + z )^2 <= 3(xx + yy + zz)
>for the given inequality it is sufficient to prove that
>3(bb + cc)^2 / (2(bb + cc) - aa) + cyclic ... <= 36(aa + bb
>+ cc)/9
>or after calculations that
>
>4aabbcc -4(bb+cc-aa)(cc+aa-bb)(aa+bb-cc)+(ab-ac)^2+
>(bc-ba)^2+(ca-cb)^2>=0
>which is true because e.g.
>a^4 >= a^4 - (bb-cc)^2 or
>a^4 >= (aa+bb-cc)(cc+aa-bb).
>
>Another nice inequality is
>1/m1 + 1/m2 + 1/m3 >= 2/R
>Does anybody has a reference for this?
>
>Best regards
>Nikos Dergiades

>From: "ndergiades"

>Reply-To: Hyacinthos@yahoogroups.com
>To: Hyacinthos@yahoogroups.com
>Subject: Re: [EMHL] S and more
>Date: Sun, 19 May 2002 09:23:55 -0000
>
>
>Sorry, a correction,
>
>
> > ******
> > If m1, m2, m3 are the medians of ABC then your inequality
> > is written
> > (bb+cc)/2m1 + (cc+aa)/2m2 + (aa+bb)/2m3 <= 6R.
> >
> > From 9RR - (aa + bb + cc) = OH^2 we get
> > aa + bb + cc <= 9RR
> > Since (x + y + z )^2 <= 3(xx + yy + zz)
> > for the given inequality it is sufficient to prove that
> > 3(bb + cc)^2 / (2(bb + cc) - aa) + cyclic ... <= 36(aa + bb
> > + cc)/9
> > or after calculations that
> >
> > 4aabbcc -4(bb+cc-aa)(cc+aa-bb)(aa+bb-cc)+(ab-ac)^2+
> > (bc-ba)^2+(ca-cb)^2>=0
>
>the correct is
>4aabbcc -4(bb+cc-aa)(cc+aa-bb)(aa+bb-cc)+
>(abb-acc)^2+(bcc-baa)^2+(caa-cbb)^2>=0
>
>
> > which is true because e.g.
> > a^4 >= a^4 - (bb-cc)^2 or
> > a^4 >= (aa+bb-cc)(cc+aa-bb).
> >
> > Another nice inequality is
> > 1/m1 + 1/m2 + 1/m3 >= 2/R
>
>Does anybody has a reference for this?
>
> Best regards
> Nikos Dergiades
>

-----Mensagem Original-----

De: Lucelindo D. Ferreira

Para: obm-l@mat.puc-rio.br

Enviada em: terça-feira, 14 de maio de 2002 10:36

Assunto: [obm-l] algebrismos

Bom dia galera. Eu queria uma mão nesse problema da olimpíada russa. Eu começei a resolver...

Prove que
a^2 + b^2 + b^2 + c^2 + a^2 + c^2 =< 6R

     2mc         2ma              2mb

Notação: a,b e c são lados do triângulo inscrito numa circunferência de raio R.

ma, mb e mc são as medianas relativas a a,b e c.

Pelo teorema isoperimétrico eu sei que

3Rsqrt3 >= a + b + c , 3R^3sqrt3 >= abc e 3R^2sqrt3 >= sqrt[(a + b + c)(b + c - a)(a + b - c)(a + c - b)]



a^2 + b^2 + b^2 + c^2 + a^2 + c^2 =< 6R

     2mc         2ma              2mb

sabendo que ma = 1/2sqrt[2(b^2 + c^2) - a^2]

mb = 1/2sqrt[2(a^2 + c^2) - b^2] e mc = 1/2sqrt[2(b^2 + a^2) - c^2]...

Empaquei aí e gostaria de sugestões para continuar a solução

Agradeço qualquer sugestão