Are homothetic additively separable preferences always a monotonic transformation of CES preferences?
In technical language, the question is the following:
Let $n>1$, and let $f:\mathbb{R}^n_{\ge 0}\rightarrow\mathbb{R}_{\ge 0}$ be continuously differentiable, concave, and homogeneous of degree one. Here, homogeneity of degree one means that for all $s\in\mathbb{R}_{\ge 0}$, and $x\in\mathbb{R}^n_{\ge 0}$, $f(sx)=sf(x)$.
And suppose that there exist continuously differentiable monotonic increasing functions $g:\mathbb{R}\rightarrow\mathbb{R}_{\ge 0}$ and $h_1,\dots,h_n:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R}$ such that for all $x\in\mathbb{R}^n_{\ge 0}$: $$f(x)=g(h_1(x_1)+\cdots+h_n(x_n)).$$
Must it be the case that there exists $a_1,\dots,a_n\in\mathbb{R}$ and $\rho,b_1,\dots,b_n\in\mathbb{R}_{\ge 0}$ such that for all $i\in\{1,\dots,n\}$, $h_i(x)=a_i+b_i \frac{x^{1-\rho}-1}{1-\rho}$? (Where, when $\rho=1$, we understand this as stating $h_i(x)=a_i+b_i\log(x)$.)
Note: This was first posted to math.se here https://math.stackexchange.com/questions/4648187/homogeneous-of-degree-one-functions-that-are-a-monotonic-transformation-of-an-ad but I think I have a higher chance of getting an answer here.
