### Autor Tema: minimum value

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06 Mayo, 2019, 09:06 am
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#### jacks

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##### minimum value
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

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##### Re: minimum value
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

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Thanks masac.