# Given $\mathbf{\alpha}$, Find the positive-entried vectors $\mathbf{x}$ maximizing $x_1^{\alpha_1}\dotsb x_n^{\alpha_n}$

A problem from an old Advanced Calculus qualifying exam:

"Choose positive real numbers $\alpha_1, \dotsc ,\alpha_n$ such that $\sum_1^n \alpha_i = 1$ and let $f:[0,\infty)^n\to \mathbb{R}:(x_1, \dotsc , x_n)\mapsto x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n}$.

Define also $C:=\{\mathbf{x}\in [0,\infty)^n \mid \sum_1^n \alpha_i x_i =1 \}$.

(a) Show that there exists $\mathbf{a}\in [0,\infty)^n$ such that $f(\mathbf{a})=\sup_{x\in C}f(x)$, and $a_i >0$ for all $i$.

(b) Find all points such that $f(\mathbf{a})=\sup_{\mathbf{x}\in C}f(\mathbf{x})$. Deduce that for any $x\in C$, $x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n}\leq 1$

(c) Deduce that for any $x\in [0, \infty)^n$, $x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n}\leq \sum_i^n \alpha_i x_i$. When does equality hold?"

I believe I've solved part (a). The relevant domain of $\mathbf{x}$ is actually compact, even though it may not look like it at first. First, if one or more $\alpha$, say $\alpha_{i_1}$ is zero, we are not concerned about $x_{i_1}$, because it does not contribute to the function, and we may regard the function as over only the other variables. Take $m=\min_{i}\{\alpha_i \mid \alpha_i \neq 0\}$, and we find that the $x_i$ that count must all be less than $1/m$. So it attains its maximum.

I have been less successful at parts (b) and (c). Here are some ideas I've been playing with: Cauchy-Schwarz says that $(\sum_1^n \alpha_i x_i)^2 \leq (\sum_{1}^n \alpha_i)^2(\sum_1^n x_i)^2$. Since $\sum_1^n \alpha_i =1$ and $\alpha_i >0$, we have that $\alpha_i^2 \leq \alpha_i$, so $\sum_1^n \alpha_i^2 \leq \sum_1^n \alpha_i$, and putting this with what we had before, we have $(\sum_1^n \alpha_i x_i)^2 \leq \sum_1^n x_i^2$. For $\mathbf{x}\in C$, $(\sum_1^n \alpha_i x_i)^2=1$, so we have that $\|x\|\geq 1\,\,\,\forall \mathbf{x}\in C$.

Also, since $\sum_1^n \alpha_i =1$, we have $\min_{i}x_i \leq x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n} \leq \max_i x_i$.

If we consider Lagrange multipliers, we have to have $\nabla f - \lambda \nabla g = 0$, for $f = x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n}$ and $g = \alpha_1x_1 + \dotsb + \alpha_nx_n$, that is, $(\alpha_1 x_1^{\alpha_1 -1}\dotsb x_n^{\alpha_n}, \dotsc, \alpha_n x_1^{\alpha_1}\dotsb x_n^{\alpha_n -1}) - \lambda \vec{\alpha} = 0$. But that's as far as I've gotten. Any ideas?

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Hint : for b) you can analize the expression $\log(x_1^{\alpha_1}x_2^{\alpha_2}\dotsb x_n^{\alpha_n})$. After some simple algebraic work you can apply Jensen inequality.