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For every $a,b,c$ non-negative real number such that:$a+b+c=1$ how to find the least value for :

$$\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$$

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7  
Do you mean to have some constraint on $a+b+c$? Otherwise $a=b=c=0$ clearly is a minimum. –  Clayton Jan 10 '13 at 15:43
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@OP do you know the Lagrange Multiplier method? –  Git Gud Jan 10 '13 at 16:38
    
@Clayton $a+b+c=1$ –  user56821 Jan 11 '13 at 3:07
    
if we will consider case $a\le b \le c$, then $(a,b,c)\approx (0,0.251454,0.748546)$. Exactly: $a=0$; $b = 1- c$; $c$ is positive root of equation $c^4+54\cdot 4c - 54 \cdot 3 = 0$. –  Oleg567 Jan 11 '13 at 4:11
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2 Answers 2

up vote 10 down vote accepted
+100

Let $\displaystyle g(x,y,z)=\frac{x}{y^3+54}+\frac{y}{z^3+54}+\frac{z}{x^3+54}$.

If we consider $g(a,b,1-(a+b))$ we can use $\partial_ag=0$ and $\partial_bg=0$ to numerically find the following critical points (up to cyclic permutation): $$\begin{array}{lll|l} \text{a} & \text{b} & \text{c} & \text{g(a,b,c)} \\ \hline 1.20836 & -0.608416 & 0.400057 & 0.0183912 \\ 0.51624 & 0.32016 & 0.1636 & 0.0185045 \\ 0.333333 & 0.333333 & 0.333333 & 0.0185058 \\ 4.69649 & -0.929554 & -2.76694 & 0.0424017\\ \end{array}$$ so in the region $a,b,1-(a+b)> 0$ it seems the minimum value is $\leq 0.0185045$. Now, for the boundaries, we use (w.l.o.g) $g(0,b,1-b)$. This has a minimum at $b=0.251454$ where $g(0,b,1-b)=0.0184826$. We also have to check So this seems to be the minimum value of $g(a,b,c)$ subject to $a+b+c=1$ and $a,b,c\geq 0$.

I tried to use Lagrange multipliers to get an exact algebraic solution, but the expressions were much too complicated (i.e. not solvable by radicals). I suspect there is no pretty closed form expression.

EDIT: By the way, here is a plot of $g(x,y,1-(x+y))$ in the region $x,y,1-(x+y)\geq 0$ (with critical points highlighted in blue and the minima highlighted in red):

Plot of g(x,y,1-(x+y)) on x,y,1-(x+y) >= 0

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After a laborious process I get the same.+1 –  daniel Jan 14 '13 at 1:44
    
@AlexanderGruber You get 0.0184826 with a = 0.748545, b = 0, c = 0.251455 which is basically what you got too I think. –  user54551 Jan 14 '13 at 22:24
    
@lip1 Indeed, that is a cyclic permutation of the boundary solution. –  Alexander Gruber Jan 16 '13 at 2:49
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Since you said non-negative, I am going to assume that zero is allowed in which case I found a minimum value of 0.0184826 achieved at $a=0, b=0.748545, c=0.251455$ up to their cyclic permutations.

After posting this I see that Oleg567 pointed to this solution already. Furthermore, if you want all $a,b,c$ to be strictly positive then it looks like the minimum is indeed at 0.0185045 although the values for $a,b,c$ where it is achieved are not unique. In addition to Alexander Gruber and daniel's answer, I found $a=0.530893, b=0.335498, c=0.133609$.

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"Furthermore, if you want all $a,b,c$ to be strictly positive then it looks like the minimum is indeed at $0.0185045$..." This is false, and in fact, you showed it was false by finding a smaller value on the boundary. Continuity tells us that in any very small ball around your boundary point, we can find a smaller value than $0.0185045$. The function actually has no global minimum on $a+b+c=1$, $a,b,c>0$. –  Eric Naslund Jan 17 '13 at 7:34
    
Yep you are right, this was done in two separate parts and of course the post was edited without any thought at all. :-) If I would have only taken five seconds longer... –  Fixed Point Jan 17 '13 at 7:54
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