Let $n \geq 2$ and $k_i, \alpha_i \in \mathbb{R}_+$ for $i \in \{1, ..., n\}$
$$ G: \mathbb{R}_+^n \to \mathbb{R}, x \mapsto G(x) = \prod_{i=1}^n x_i^{\alpha_i} - \sum_{i=1}^n k_i x_i $$ If $\sum_{i=1}^n \alpha_i > 1$, show $G$ neither has a global nor a local extremum.
I don't have a clue on how to prove this. Can anyone help out?
$G(x_1,...x_n)=\prod_{i=1}^n x_i^{\alpha_i} - \sum_{i=1}^n k_i x_i$, so $G_{x_j}=\alpha_jx_j^{\alpha_j-1}\prod_{i\ne j}x_i^{\alpha_i}-k_j$.
If G has a local extremum, then at the point where the extreme occurs, $G_{x_j}=0 \;\forall j$. Therefore $\alpha_jx_j^{\alpha_j-1}\prod_{i\ne j}x_i^{\alpha_i}=k_j$ or $G(x_1,...x_n)\alpha_j=k_jx_j$, which implies $G_{x_j}=\dfrac{k_j}{\alpha_j}$. Clearly, this expression is never 0, and hence G doesn't have a local extremum. Since G doesn't have a local extremum, it doesn't have a global extremum either. In fact, $G_{x_j}\gt 0 \; \forall j$ tells us that if we fix all the other variables and only let one variable, say $x_j$, vary, then the value of G will increase monotonically in $x_j$.