Autor Tema: minimum value

0 Usuarios y 1 Visitante están viendo este tema.

06 Mayo, 2019, 09:06 am
Leído 575 veces

jacks

  • $$\Large \color{red}\pi\,\pi\,\pi\,\pi\,\pi$$
  • Mensajes: 672
  • País: in
  • Karma: +0/-0
  • Sexo: Masculino
If \( a,b,c,d> 0. \) Then minimum value of \( \displaystyle a^5+3\sqrt{3}\;b^5+\sqrt{3}c^5+d^5-15abcd \)

06 Mayo, 2019, 12:13 pm
Respuesta #1

Masacroso

  • Moderador Global
  • Mensajes: 2,227
  • País: es
  • Karma: +0/-0
If \( a,b,c,d> 0. \) Then minimum value of \( \displaystyle a^5+3\sqrt{3}\;b^5+\sqrt{3}c^5+d^5-15abcd \)

this is wrong, forget it
If \( a,b,c \) and \( d \) are real then there is no minimum value, just choosing \( a=b=c=d \) then we have the polynomial \( p(a)=((2+4\sqrt 3)a-15) a^4 \) and \( p(0)=0 \).

So I guess that \( a,b,c,d\in\Bbb N_{>0} \). Now note that \( 3\sqrt 3>\sqrt 3>1 \) and so we must have that \( a,d\ge c\ge b \) to minimize the polynomial. Then it seems that when \( a\ge 2 \) then the value of the polynomial is positive, that is, we have \( 32+d^5+3\sqrt 3b^5+\sqrt 3 c^5-30bcd>0 \) for \( b,c,d\in\{1,2\} \) and \( a=2 \) (for \( a>2 \) the value of the polynomial is bigger), then probably (I dont have a proof by now) the minimum happen when \( a=b=c=d=1 \).
[cerrar]

For real valued \( a,b,c,d>0 \) we can find the gradient of the polynomial \( p

(a,b,c,d):=a^5+3\sqrt 3 b^5+\sqrt 3 c^5+d^5-15abcd \) to find a local minimum for positive values, that is:

\( \displaystyle \nabla p(a,b,c,d)=(5a^4-15bcd, 15\sqrt 3 b^4-15acd, 5\sqrt 3 c^4-15abd, 5d^4-

15 abc) \)

Then the critical points of the polynomial are at the points defined by the equations

\( \displaystyle a^4=3bcd,\,\sqrt 3 b^4=acd,\, c^4=\sqrt 3 abd,\, d^4=3abc,\, a,b,c,d>0\tag1

 \)

Hence \( a=d \), from the previous first and last equation, and \( 3^{1/5}b=c \) from the other two equations, so the equations now become

\( \displaystyle a^3=3^{6/5}b^2,\, 3^{3/10} b^3=a^2,\, a,b>0\iff ba=3^{9/10}\tag2 \)

Then, if there is no mistake somewhere, for any chosen \( a>0 \) there is a critical point for \( d=a,\, b=3^{9/10}a^{-1},\, c=3^{11/10}a^{-1} \). In this case we have that

\( \displaystyle p(a,b,c,d)=2a^5+(3^6+3^{8/5})a^{-5}-15\cdot 9\tag3 \)

so the anser would be the minimization of the last expression, that is achieved when \( 2-Kx^{-2}=0\iff x=\sqrt{K/2} \) for \( x:=a^5 \) and \( K:=3^6+3^{8/5} \).



Alternatively, from (2) we can get the identity \( a^5=3^{3/2}b^5,\, a,b>0\iff a=3^{3/10}b \). Hence the polynomial becomes \( (4a-5\cdot 3^{3/5})a^4 \) that is minimized when \( 20a^3(a-3^{3/5})=0 \) and \( a>0 \), so the minimum is at \( a=3^{3/5},\, b=3^{3/10},\, c=3^{-1/10},\, d=3^{3/5} \).

06 Mayo, 2019, 05:26 pm
Respuesta #2

jacks

  • $$\Large \color{red}\pi\,\pi\,\pi\,\pi\,\pi$$
  • Mensajes: 672
  • País: in
  • Karma: +0/-0
  • Sexo: Masculino
Thanks masac.