Maximum of a rational homogeneous function

Question

I have rational functions that I need to maximize over $\mathbb{R}_+^n \setminus {0}$.

Here is an example for $n=4$ : $$ f(a,b,c,d) = \frac{a(d + a + b) + b(a + b + c) + c(b + c + d) + d(c + d + a)}{{(a + b + c + d)}^2} $$

I remark that $f$ is a quotient of $2$-homogeneous polynomials, so $\forall \lambda, \forall v \in R^4, f(\lambda v) = f(v)$, and therefore it is enough to maximize $f$ on the (positive part of the) $3$-sphere.

I wish to prove that the maximum is in $f(1,1,1,1) = \frac{3}{4}$.

Does anyone see a way of proving this "simply" (say, not using the Hessian) ?

Many thanks !

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EDIT

As you pointed out, my claim was wrong. As it happens, I oversimplified my problem and failed to see it. Thank you all for pointing me wrong.

For the record, the function I really need to maximize when $n=4$ would be $$ f(a,b,c,d) = \frac{a(d + a + b) + b(a + b + c) + c(b + c + d) + d(c + d + a)}{(a + b + c + d) \cdot max(a+b, b+c, c+d, d+a)} $$

For this one all my numeric tests (besides the geometric intuition behind this question) have concluded that the maximum is in $(1,1,1,1)$.

Answer 1

The claim about the maximum over $\mathbb{R}_+^4 \setminus {0}$ is not true, since $f(1,1,1,5/4)=217/289>3/4$. In fact, $$ f(1,1,1,d)=\frac{d^2+4d+7}{d^2+6d+9}. $$

Answer 2

The general problem is probably very hard.

In this example, it is easier to maximize when $a+b+c+d=1$ rather than $a^2+b^2+c^2+d^2=1.$ When $a+b+c+d=1,$ then:

$$f(a,b,c,d)=a(1-c)+b(1-d)+c(1-a)+d(1-c)= 1-2(ac+bd).$$

This is maximized, for positive $a,b,c,d$ when $ac+bd$ is minimized. We can't minimize $ac+bd$ to $0.$ But given any$0<\epsilon<\frac 13$, we can get $ac+bd<\epsilon$ by taking $$(a,b,c,d)=\left(\epsilon,\epsilon,\epsilon,1-3\epsilon\right)$$ Then $ac+bd=\epsilon - 2\epsilon^2<\epsilon.$

So $f(a,b,c,d)<1$ on our domain, and can be made arbitrarily close to $1$, but never equal to $1$ since $a,b,c,d>0.$

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If you wanted to minimized $f$, on the other hand, we again restrict $a+b+c+d=1$ then take $a+c=\alpha$ and $b+d=1-\alpha.$ Then AM/GM gives $$ac\leq \frac{(a+c)^2}{4}=\frac{\alpha^2}{4},bd\leq\frac{(1-\alpha)^2}4$$ with equality when $a=c=\frac{\alpha}{2},b=d=\frac{1-\alpha}{2}.$ Then we get the maximum value of $ac+bd$ is the maximum value of

$$\frac{\alpha^2}{4}+\frac{(1-\alpha)^2}{4}=\frac{2\alpha^2-2\alpha+1}{4}=\frac{(2\alpha-1)^2+1}{8}$$

This takes the maximum value when $\alpha=0,1$ which would put us in $abcd=0$ territory. But if $\alpha=\epsilon$ then $(a,b,c,d)=\left(\frac\epsilon 2,\frac{1-\epsilon}2,\frac \epsilon 2,\frac {1-\epsilon}2\right)$ gives us $$f(a,b,c,d)=1-2(ac+bd) = 1-\frac{(1-2\epsilon)^2+1}{4}=\frac{1}{2}+\epsilon(1-\epsilon)$$

So the infimum of $f(a,b,c,d)$ is $\frac{1}{2}$ and the supremum is $1,$ but it never equals either when $a,b,c,d$ are positive.

Answer 3

$$(a+b+c+d)^2-\sum_{cyc}(a^2+ab+ad)=(a+b+c+d)^2-\sum_{cyc}(a^2+2ab)=2(ac+bd)>0$$ and for $a=b\rightarrow0^+$ we get $2(ac+bd)\rightarrow0^+.$

Thus, the maximum does not exist and $$\sup f=1.$$