The concavity of $x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ in $\mathbb{R}^n_+$ with exponent less than $1$ Let $f(x) : \mathbb{R}^n_+ \rightarrow \mathbb{R}_+$ be defined as 
$$f(x) = x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n},$$
where $\sum\limits_{i=1}^na_i=1$, $a_i \geq 0$.
I am wondering if it can be proved that it is concave. I initially thought it was easy, but I tried computing Hessian as well as using the definition without any progress.
 A: $\def\T{^{\mathrm{T}}}$Without loss of generality, assume that $0 < a_k < 1$ for all $k$. Since$$
\frac{∂^2 f}{∂x_k ∂x_j}(x) = \begin{cases}\dfrac{a_k (a_k - 1)}{x_k^2} f(x); & k = j \\ \dfrac{a_k a_j}{x_k x_j} f(x); & k ≠ j\end{cases},$$
then$$
(Hf)(x) = f(x) \begin{pmatrix}
\dfrac{a_1 (a_1 - 1)}{x_1^2} & \dfrac{a_1 a_2}{x_1 x_2} & \cdots & \dfrac{a_1 a_n}{x_1 x_n}\\
\dfrac{a_2 a_1}{x_2 x_1} & \dfrac{a_2 (a_2 - 1)}{x_2^2} & \cdots & \dfrac{a_2 a_n}{x_2 x_n}\\
\vdots & \vdots & \ddots & \vdots\\
\dfrac{a_n a_1}{x_n x_1} & \dfrac{a_n a_2}{x_n x_2} & \cdots & \dfrac{a_n (a_n - 1)}{x_n^2}
\end{pmatrix}.
$$
Denote $A(x) = \dfrac{1}{f(x)} (Hf)(x)$, then it suffices to prove that $A(x) \leqslant 0$. For any $y = (y_1, \cdots, y_n)\T \in \mathbb{R}^n$, define $t_k = \dfrac{y_k}{x_k}$, then\begin{align*}
y\T A(x) y &= \sum_{k = 1}^n \frac{a_k (a_k - 1)}{x_k^2} · y_k^2 + \sum_{k ≠ j} \frac{a_k a_j}{x_k x_j} · y_k y_j = \sum_{k = 1}^n a_k (a_k - 1) t_k^2 + \sum_{k ≠ j} a_k a_j t_k t_j\\
&= \left( \sum_{k = 1}^n a_k^2 t_k^2 + \sum_{k ≠ j} a_k a_j t_k t_j \right) - \sum_{k = 1}^n a_k t_k^2 = \left( \sum_{k = 1}^n a_k t_k \right)^2 - \sum_{k = 1}^n a_k t_k^2 \leqslant 0,
\end{align*}
where the last step uses Jensen's inequality. Therefore $A(x) \leqslant 0$, then $f$ is concave.
