# Optimizing elementary symmetric polynomial on the unit sphere

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.

• There are two critical points where $x_1=x_2=x_3=x_4$. ${}\qquad{}$ – Michael Hardy Jun 29 '14 at 3:36
• oh: You meant in the positive orthant. ${}\qquad{}$ – Michael Hardy Jun 29 '14 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 '14 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 '14 at 3:49

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.

• I like this better than my answer, very elegant. – Sergio Parreiras Jun 29 '14 at 3:53
• Thanks! That's very nice of you! :-) – Adam Hughes Jun 29 '14 at 3:58
• That's amazing! – Mark Jun 29 '14 at 4:08
• 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 '14 at 17:17

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, $$\sqrt{(c+d)xy}$$.

• What does quasi concave mean? – Mark Jun 29 '14 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 '14 at 20:26
• Is there an easy way to see that $f(x_1,x_2)$ is quasi-concave? – Mark Jun 29 '14 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 '14 at 22:08
• 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 '14 at 22:23