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I'd like to optimize $x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4$ on the unit 4-sphere. I'm thinking I should do lagrangian optimization, but I'm having trouble solving the resulting system.

We may assume that the maximum is attained when $x_i \gt 0$ for all $i$.

I happened to "guess" one of the critical points which is at $x_1 = x_2 = x_3 = x_4$, so any argument which shows there is only one critical point will suffice.

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There are two critical points where $x_1=x_2=x_3=x_4$. ${}\qquad{}$ –  Michael Hardy Jun 29 at 3:36
    
oh: You meant in the positive orthant. ${}\qquad{}$ –  Michael Hardy Jun 29 at 3:37
    
"the maximum is attained when $x_i>0$ for all $i$" should not be taken to mean "the maximum is attained ONLY when $x_i>0$ for all $i$". ${}\qquad{}$ –  Michael Hardy Jun 29 at 3:39
    
Yes, that is what I meant, but I didn't know how to word it other than how I did. –  Mark Jun 29 at 3:49
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2 Answers 2

up vote 1 down vote accepted

Your sum is a sum of geometric means of $x_1^3, x_2^3, x_3^3,x_4^3$ you know each is less than the arithmetic means so the max, $M$, satisfies

$$M\le x_1^3+x_2^3+x_3^3+x_4^3=f(x_1,x_2,x_3,x_4)$$

On the sphere, $f$ has a maximum at $(x_1^2,x_2^2,x_3^2,x_4^2)=\lambda(x_1,x_2,x_3,x_4)$ implying $x_i=x_j$ for all $i,j$. Since the AM is maximized there and GM=AM at this point, you know that there is no means to get a larger value than there.

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I like this better than my answer, very elegant. –  Sergio Parreiras Jun 29 at 3:53
    
Thanks! That's very nice of you! :-) –  Adam Hughes Jun 29 at 3:58
    
That's amazing! –  Mark Jun 29 at 4:08
    
If you like it, I believe it's polite to accept the answer. ;-) –  Adam Hughes Jun 29 at 7:08
    
Just waiting to see if someone will post a more "standard" solution. I can't hope to get this creative on the exam. –  Mark Jun 29 at 17:17
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My suggestion is a bit different:

  1. Show that the maximum is not going to change if you expand the set to be the intersection of the positive orthant and the unit ball ( a convex set )
  2. Show that if a solution $(a,b,c,d)$ is such that $a\neq b$ the we have other solution(s) that can be obtained by permutation of the entries, for example, $(b,a,c,d)$.
  3. Show that for $a,b,c,d>0$ we have $f(a,b,c,d)=f(b,a,c,d)$ and $f(\frac 12 (a+b), \frac 12 (a+b),c,d)>f(a,b,c,d)$. For this part think of $f$ as a function of only $x_1$ and $x_2$ and hold $x_3=c$ and $x_4=d$. Since $f(x_1,x_2,c,d)=(c+d)(x_1 \cdot x_2) + cd(x_1+x_2)$ is strictly quasi-concave for $c+d>0$ we are done.

An easy way to see $f(x,y)=(c+d)xy+cd(x+y)$ is stricly concave in the positive orthant is to see it is the sum of a stricly concave function with a linear one. Too see $(c+d)xy$ is stricly concave notice it is an increasing transformation of a strictly concave function, $\sqrt{(c+d)xy}$.


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What does quasi concave mean? –  Mark Jun 29 at 4:05
    
@Mark: quasi-concave means the set $\{ x : f(x)\ge a \}$ is convex for any $a$ but there are lots of equivalent but different definitions. For example: $xy$ is not concave but it is quasi-concave. –  Sergio Parreiras Jun 29 at 20:26
    
Is there an easy way to see that $f(x_1,x_2)$ is quasi-concave? –  Mark Jun 29 at 21:57
    
And could you clarify the strict part? If the function is quasi-concave, I see that $f((a+b)/2, (a+b)/2)$ has to be greater than or equal to $f(a,b)=f(b,a)$, but why would it have to be strictly greater? I guess I could prove it by parameterizing the line between $(a,b)$ and $(b,a)$ and doing a 1-dimensional optimization, but is there an easy way to see it? –  Mark Jun 29 at 22:08
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Maybe a simpler argument avoiding quasi-convexity which I'm not so familiar with: Parameterizing the line between $(a,b)$ and $(b,a)$ and evaluating the derivative at 0 shows a positive derivative which contradicts maximality at $(a,b)$ –  Mark Jun 29 at 22:23
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