Let $S\in\mathbb{R}^n$ be a convex set containing the zero vector $(0,...,0)$, and assume that for all $i=1,...,n$ $$\max_{x\in S\cap\mathbb{R}^{n}_{+}}x_{i}=1.$$ To prove is then that $$\arg\max_{x\in S\cap\mathbb{R}^{n}_{+}}\prod_{i=1}^{n}x_{i}\geq(1/n,...,1/n)$$ I suspect that the argument must involve the fact that $(1/n,...,1/n)$ is the maximizer of the product function in the unit simplex, to which we then apply a concave transformation, but this is just a guess. Any help would be much appreciated.
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I am not sure I understand what you mean. But without loss of generality we may assume that the unit simplex ${x\in \mathbb{R}^{n}{+}\mid x{1}+\ldots+ x_{n}=1}$ is in $S\cap\mathbb{R}^{n}_{+}$ because the maximizer of the product function is never in the boundary of S where all components can be increased. Does that change things? – Orosius Mar 17 '19 at 18:01
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Guideline: (1) prove that maximizing $x_1 \dots x_n$ is the same as minimizing $-\log(x_1) - \dots - \log(x_n)$; (2) define $C = S \cap \mathbb{R}^n_{+}$ and use the optimality condition for convex programming of minimizing $f(x)$ s.t. $x \in C$, which is $\nabla f(x^)^T (y - x^) \geq 0$ for all $y \in C$. From there the proof is quite easy. – Alex Shtoff Mar 18 '19 at 11:37
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Thanks Alex, very helpful. – Orosius Mar 18 '19 at 14:43
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did you manage to prove it? – Alex Shtoff Mar 19 '19 at 13:44
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I did, thanks again! – Orosius Mar 20 '19 at 10:55