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Find min & max of $a(b-c)^n+b(c-a)^n+c(a-b)^n$ where $a + b+ c =1;\ a,b,c\ge0; \ n \in N$

I am really stuck, I don't remember where I read this problem.

Xeing
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  • Do you know the Lagrange multipliers? – Sebastien B Feb 19 '13 at 12:14
  • Yes, a little, but can you prove for case $n=3$ without use Lagrange multipliers ? – Xeing Feb 19 '13 at 12:32
  • @D3r0X4 - what's wrong with Lagrange multipliers? – Nathaniel Bubis Feb 19 '13 at 12:36
  • You can probably do something without Lagrange multipliers, by considering the following equivalent problem: find the $\min$ and $\max$ of $a(2b+a-1)^n+b(1-2a-b)^n+(1-a-b)(a-b)^n$, for $a,b,1-a-b\geq 0$. But it's probably not very nice. – Sebastien B Feb 19 '13 at 12:41
  • @nbubis: can you show me the way to solve it using Lagrange multipliers. I still not familiar with it (I just read it yesterday) – Xeing Feb 19 '13 at 12:52
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    The expression for the max value does not seem nice for general $n$. If we write $[m, M]$ to mean $m$ is min and $M$ is max, then $n=2$ gives $[0, \frac{1}{4}]$, $n=3$ gives $[-\frac{1}{6\sqrt{3}}, \frac{1}{6\sqrt{3}}]$, $n=4$ gives $[0, \frac{1}{12}]$, $n=5$ gives $[-\frac{1}{\sqrt[4]{5}}, \frac{1}{\sqrt[4]{5}}]$, $n=6$ gives $[0, \frac{1}{135}(4\sqrt{10}-5)]$. Well actually it isn't that bad. I'll see if I can find a general formula. – Ivan Loh Feb 20 '13 at 01:28
  • Since I can't edit my above comment anymore: Typo: It should be $n=5$ is $[-\frac{1}{10\sqrt[4]{5}}, \frac{1}{10\sqrt[4]{5}}]$ – Ivan Loh Feb 20 '13 at 01:58
  • I take my comment back. The expressions get worse and worse. The maximum for $n=7$ is probably: $$\frac{\sqrt{\frac{\left(-585893+\frac{52166103199}{\sqrt[3]{\frac{1}{7}(-4495722125892077+63452210761728 i \sqrt{17})}}+7^{2/3} \sqrt[3]{-4495722125892077+63452210761728 i \sqrt{17}}\right)}{7}}}{1512}$$ (Not proven but probably true) (The numbers come from the 1 variable maximisation of a polynomial of degree $n$, which I did by hand up to $n=6$. For $n=7$, the expression comes from Wolfram Alpha.) – Ivan Loh Feb 20 '13 at 02:44
  • Maybe you are supposed to find the minimum and maximum of the expression over all $n$, and not for specific $n$? – Ivan Loh Feb 20 '13 at 02:48

1 Answers1

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Without using Lagrange for $n=3$:

Substitute $c = 1 - a -b$ to get: $$y = (b - a) (2 a + b - 1) (a + 2 b -1)$$ The condition $a,b,c>0$, leads to us looking for a extrema on the triangle $0<a<1, a+b < 1$. Solving for $y_a = 0, y_b = 0$, you get that $a=b=1/3$ is a solution but it's only an inflection point. So, the minima and maxima must lie on the borders of the triangle:

For $b=0$: You get $y = -(-1 + a) a (-1 + 2 a)$. Derive and compare to $0$ to get: $$a = \frac{1}{2}\pm\frac{\sqrt{3}}{6} \to y = \pm\frac{1}{6\sqrt{3}}$$

For $b=0$: Due to symmetry, you get the same numbers with the positions reversed.

For $a+b=1$: This is really $c=0$, so again due to symmetry, the same results hold. All in all the maximum and minimum values are given by: $$y = \pm\frac{1}{6\sqrt{3}} \sim \pm0.096225$$                                              enter image description here

Nathaniel Bubis
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  • I have a question after use Lagrange I get: $\lambda = (b-c)(b^2+c^2+bc-3a^2) = (c-a)(c^2+a^2+ac-3b^2) = (a-b)(a^2+b^2+ab-3c^2)$

    and $a+b+c=1$

    $\implies a=b=c=\frac 13 \implies y=0 \ ?$ Where is the wrong ?

     – Xeing Feb 19 '13 at 13:44
  • Lagrange does not guarantee that the points you find will indeed be local extrema. As I mentioned, the point you found is an inflection point. – Nathaniel Bubis Feb 19 '13 at 13:52