Proving a function is monotone 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?
 A: 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.
A: Write 
$$p(\alpha)= \sum_{i=1}^n \left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right)^{\alpha}$$
Then you have 
$$p'(\alpha)= \sum_{i=1}^n \ln \left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right) \times \left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right)^{\alpha}$$
i.e. $$p'(\alpha)  =  \sum_{i=1}^n \left( \ln(u_i) - \frac{1}{n} \sum_{k=1}^n \ln(u_k)\right)\times \left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right)^{\alpha}  $$
You deduce first that 
$$p'(0)  =  \sum_{i=1}^n \left( \ln(u_i) - \frac{1}{n} \sum_{k=1}^n \ln(u_k)\right)= 0$$
Moreover, you get 
$$p''(\alpha) = \sum_{i=1}^n \ln ^2\left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right) \times \left( \frac{u_i}{\left(\prod_{k=1}^n u_k \right)^{\frac{1}{n}}}\right)^{\alpha} > 0$$
So $p'$ is increasing and vanishes when $\alpha = 0$, so $p'$ is positive for $\alpha >0$. You deduce that $p$ is increasing.
