$\prod_{i=1}^n x_i^{\alpha_i}$ is neither convex nor concave when $\sum_i \alpha_i > 1$ 
Let $n\geq 2$ and $\displaystyle f:(x_1,\ldots,x_n) \mapsto \prod_{i=1}^n x_i^{\alpha_i}\quad$ be defined for positive $x_i$
Suppose $\sum_{i=1}^n \alpha_i > 1$. Prove that $f$ is neither concave nor convex.

The Hessian matrix of $f$ is easy to compute. It suffices to prove that it is neither positive semi-definite, neither negative semi-definite at some point.
Let $H(f)(x)$ denote the Hessian matrix of $f$ at $x=(x_1,\ldots,x_n)$. It suffices to find some $x$ such that $\det H(f)(x)<0$ or one may look for $u$ and $v$ such that $u^T H(f)(x)u> 0$ and $v^T H(f)(x)v< 0$.
I've noted that $$\frac{\partial^2 f}{\partial^2 x_i}(x) = \alpha_i(\alpha_i-1)\frac{f(x)}{x_i^2}$$
and $$\frac{\partial^2 f}{\partial x_j \partial x_i}(x) = \alpha_i \alpha_j\frac{f(x)}{x_i x_j}$$
I think the approach with the determinant is the most tractable.  $x=(\alpha_1,\ldots, \alpha_n)$ seems like a promising point, although I'm not through with the computations.
EDIT: it seems that the approach with the determinant would require that $n$ is even...
EDIT 2: Provided my computation is correct, at $x=(\alpha_1,\ldots, \alpha_n)$, $$\det H(f)(x)=(f(x))^n\frac{(-1)^{n-1}}{\prod_{i=1}^{n}\alpha_i} \left[\left(\sum_{i=1}^n \alpha_i \right) -1 \right]$$
This solves the problem, with $n$ even. It remains to take care of odd $n$.
 A: This is a partial answer, proving that the determinant of the Hessian is of constant sign.
Looking at specific cases
Case $n = 2$
At least a partial solution for $n = 2$. I haven't try to extend it for $n>2$.
For $x=(1,1)$ you have $\det H(f)(1,1)=\alpha_1 \alpha_2 (1-\sum \alpha_i)<0$.
And it could be interesting to still look at the value of $\det H(f)(x)$ for $x = (1, \dots,1)$ and $n>2$.
Case $n = 3$
Take $f(x_1,x_2,x_3)=x_1 x_2 x_3$. You have in that case $\det H(f)(x)=2x_1 x_2 x_3 >0$, proving that determinant of the Hessian is always positive in that case.
The general case
Denoting $\alpha^\prime_i = \frac{\alpha_i}{x_i}$ and $\beta_i = \frac{1}{x_i}$, we have:
$$
\begin{aligned}\det H(f)(x)&= (f(x))^n
\begin{vmatrix}
(\alpha^\prime_1 - \beta_1) \alpha^\prime_1 &  \alpha^\prime_1 \alpha^\prime_2 & \dots & \alpha^\prime_1 \alpha^\prime_n\\
\alpha^\prime_1 \alpha^\prime_2 &  (\alpha^\prime_2 - \beta_2) \alpha^\prime_2 & \dots & \alpha^\prime_2 \alpha^\prime_n\\
\vdots &  \ddots &  & \vdots\\
\alpha^\prime_1 \alpha^\prime_n &  \dots & \alpha^\prime_{n-1} \alpha^\prime_n & (\alpha^\prime_n - \beta_n) \alpha^\prime_n\\
\end{vmatrix}\\
&=(f(x))^n (\alpha^\prime_1 \dots \alpha^\prime_n)^2 \begin{vmatrix}
1-\frac{1}{\alpha_1} &  1 & \dots & 1\\
1 &  1-\frac{1}{\alpha_2} & \dots & 1\\
\vdots &  \ddots &  & \vdots\\
1 &  \dots & 1 & 1-\frac{1}{\alpha_n}\\
\end{vmatrix}
\end{aligned}$$ proving that the sign of the determinant of the Hessian is independent from the point $x$. And according to your computation for $(x_1, \dots,x_n)=(\alpha_1, \dots,\alpha_n)$ this sign is positive for $n$ odd and negative for $n$ even.
A: If $f$ was convex or concave, its graph would sit on a single side of its tangent plane
$$\varphi(x_1, \dots, x_n)  = 0$$  with
$$\varphi(x_1, \dots, x_n)=\alpha_1(x_1-1) + \dots + \alpha_n (x_n-1)$$ at point $U=(1, \dots, 1)$
For the origin $O$ we have
$$\varphi(O)-f(O)=- (\alpha_1 + \dots + \alpha_n) < -1 \le 0$$
Now consider $i_0$ for which $\alpha_{i_0} > 0$ and a point $P_t$ whose coordinates are all vanishing except the $i_0$-one which is equal to $t>0$. You have $\varphi(P_t)-f(P_t)= \alpha_{i_0}t - \sum_i \alpha_i > 0$ for $t>\frac{\sum_i \alpha_i}{\alpha_{i_0}}$.
Hence the graph of $f$ is not on a single side of its tangent plane at $U$ proving that $f$ is not convex nor concave.
