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Given function $$f:\mathbb{R}^n_{+} \rightarrow \mathbb{R}, \ f(x) = \sum_{k=1}^K c_k x_1^{a_{1k}} x_2^{a_{2k}} \cdots x_n^{a_{nk}}$$

Show that for any $x, y \in \text{dom} \ f, \theta \in [0,1]$, and let $z : z_i = x_i^{\theta} y_i^{1-\theta}$, the following holds $$ f(z) \leq f(x)^{\theta} f(y)^{1-\theta}$$

I know I need to to make the transformation $$ w_i = \log (x_i), \\ \ f(x) = \sum_{i=1}^K c_k (e^{w_1})^{a_{1k}} (e^{w_2})^{a_{2k}} \cdots (e^{w_n})^{a_{nk}} \\ = \sum_{i=1}^K e^{a_k^T w + b_k} \quad \text{where $b_k = \log c_k$} $$ Then $f(x)$ is convex. But I'm not sure how to proceed from here.

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In general, your function $f$ is nonconvex! For example, on $\mathbb R_+^2$ consider $f :(x_1, y_2) \mapsto x_1x_2$. However, the inequality your trying to proof holds for arbitrary $f$ defined as you did. When such an inequality holds, we say $f$ is multiplicatively convex.

So then, let's show that your $f$ is multiplicatively convex. Indeed, $$f(x) = \sum_k c_kx_1^{a_{1,k}}\ldots x_n^{a_{n,k}} = \sum_k e^{a_k^Tw(x) + b_k}, $$ where $w(x) := (\log x_1, \ldots, \log x_n)$, and $b_k := \log c_k$.

Now, $w(z) = (\log(x_1^\theta y_1^{1-\theta}), \ldots, \log(x_n^\theta y_n^{1-\theta})) = \ldots = \theta w(x) + (1-\theta) w(y)$. Thus $$ \begin{split} f(z) &= \sum_k e^{a_k^T(\theta w(x) + (1-\theta) w(y)) + b_k} = \sum_k e^{\theta (a_k^Tw(x) + b_k)}e^{(1-\theta) (a_k^Ty(x) + b_k)}\\ &= \sum_k e^{\theta (a_k^Tw(x) + b_k)}\sum_k e^{(1-\theta) (a_k^Ty(x) + b_k)} \le \left(\sum_k e^{a_k^Tw(x) + b_k}\right)^\theta\left(\sum_k e^{a_k^Ty(x) + b_k}\right)^{1-\theta}\\ &\hspace{7.6cm}= f(x)^\theta f(y)^{1-\theta}, \end{split} $$ where the last inequality holds because $\theta, 1-\theta \in [0,1]$. Thus $f$ is multiplicatively convex.

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