Let $n\in \mathbb{N}$, $u_1,u_2,\ldots ,u_n>0$ and I want to prove that the function $$p(\alpha)=\frac{\sum_{i=1}^n u_i^\alpha}{\left( \prod_{i=1}^n u_i^\alpha \right)^{1/n}}$$ is monotone in relation to $\alpha>0$.
The derivative is $$p'(\alpha)=\frac{\sum_{i=1}^n u_i^\alpha \ln u_i \left(\prod_{i=1}^n u_i^\alpha \right)^{1/n}-\frac1n\left(\sum_{i=1}^nu_i^\alpha\right)\left(\sum_{i=1}^n\ln u_i\right)\left(\prod_{i=1}^n u_i^\alpha \right)^{1/n} }{\left(\prod_{i=1}^n u_i^\alpha \right)^{2/n}}$$ but there doesn't seem to be a reason why it should always be positive/negative. I also tried taking derivative of $\ln p(\alpha)$. Any ideas?
First note that the function $p$ can be written as sum of exponential functions as follows: \begin{align} p(\alpha) &=\frac{\sum_{i=1}^n u_i^\alpha}{\left( \prod_{j=1}^n u_j^\alpha \right)^{1/n}}\\ &=\sum_{i=1}^n\left(\frac{u_i}{\prod_{j=1}^n u_j^{1/n}}\right)^\alpha\\ &=\sum_{i=1}^ne^{r_i\alpha} \end{align} where \begin{align} r_i &=\ln\left(\frac{u_i}{\prod_{j=1}^n u_j^{1/n}}\right)\\ &=\ln(u_i)-\frac 1n\sum_{j=1}^n \ln(u_j) \end{align} Consequently, we have \begin{align} p'(\alpha)&=\sum_{i=1}^ne^{r_i\alpha}r_i& p'(0)&=\sum_{i=1}^nr_i=0 \end{align} Since for every $\alpha$ we have $$p''(\alpha)=\sum_{i=1}^ne^{r_i\alpha}r_i^2\geq 0$$ we get $p'(\alpha)\geq p'(0)=0$ for every $\alpha>0$, hence $p$ is a non-decreasing function.