Optimizing elementary symmetric polynomial on the unit sphere

Question

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.

Answer 1

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.

Answer 2

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 quasi-concave in the positive orthant is to see it is the sum of a stricly quasi-concave function with a linear one. Too see $(c+d)xy$ is stricly quasi-concave notice it is an increasing transformation of a strictly concave function, $\sqrt{(c+d)xy}$.

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