I want to show that
(1) the value of the following formula increases as $m$ increases and
(2) the limit or the maximum value is equal to the maximum among $x_i$.
Any thoughts or suggestions are greatly appreciated!
$$ f = \frac{{\sum\limits_{i = 1}^n {{a_i}x_i^m} }}{{\sum\limits_{i = 1}^n {{a_i}x_i^{m - 1}} }},{a_i} > 0,{x_i} \ge 1, m \ge 1 $$
Proof of the second part
Following @sku's suggestion: suppose $x_k$ is the maximum among $x_i$, then divide the numerator and denominator by $a_k x_k^{m-1}$, the problem can be simplified as follows:
$$ \begin{array}{l} {b_i} = \frac{{{a_i}}}{{{a_k}}} > 0,\\ {y_i} = \frac{{{x_i}}}{{{x_k}}} \in \left( {0,1} \right],y_i^{m - 1} = \frac{{x_i^{m - 1}}}{{x_k^{m - 1}}} \in \left( {0,1} \right],\\ {w_i} = {b_i}y_i^{m - 1} \in \left( {0,{b_i}} \right] \end{array} $$
The equlity holds when $i=k$. Now $f$ becomes:
$$ f = \frac{{\sum\limits_{i = 1}^n {{b_i}y_i^{m - 1}{x_i}} }}{{\sum\limits_{i = 1}^n {{b_i}y_i^{m - 1}} }} = \frac{{\sum\limits_{i = 1}^n {{w_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{w_i}} }} $$
Let $w_{i}^{\prime}=\frac{w_{i}}{\sum_{j=1}^{n} w_{j}}$, we get
$$ f = \sum_{i=1}^{n} w^{'}_{i} x_i $$
$f \le x_k$ because
$$ {x_k} - \sum\limits_{i = 1}^n {{w^{'}_i x_i}} = \sum\limits_{i = 1}^n {w_i^{'}} \left( {{x_k} - {x_i}} \right) \ge 0 $$ The equality holds when $w^{'}_k=1$ and $w^{'}_i=0,~\forall i \ne k$. This proves the second part.
Proof of the first part
Following a friend's suggestion: we can prove the first part by treating $f$ as a function of ${m}$ and then showing that $f^{'}({m})>0$.
Let $g=\sum\limits_{i = 1}^n {{a_j}y_j^{m-1}}$, then taking the derivative of $f({m})$ we get
$$ \begin{align} {f^{'}}\left( {m} \right) &= \sum\limits_{j = 1}^n {\frac{{{a_j}y_j^{m-1}{x_j} \cdot \sum\limits_{i = 1}^n {{a_i}y_i^{m-1}\left( {\ln {y_j} - \ln {y_i}} \right)} }}{{{g^2}}}} \\ & = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\frac{{{a_i}{a_j}y_i^my_j^{m-1}}}{{{g^2}}}\left[ {{x_j}\left( {\ln {y_j} - \ln {y_i}} \right)} \right]} } \\ & = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\frac{{{a_i}{a_j}y_i^my_j^{m-1}}}{{{g^2}}}\left[ {{x_j}\left( {\ln \frac{{{x_j}}}{{{x_k}}} - \ln \frac{{{x_j}}}{{{x_k}}}} \right)} \right]} } \\ & = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\frac{{{a_i}{a_j}y_i^my_j^{m-1}}}{{{g^2}}}\left[ {{x_j}\left( {\ln {x_j} - \ln {x_i}} \right)} \right]} } \end{align} $$
Let $0 \le \varepsilon \le \frac{{{a_i}{a_j}y_i^my_j^{m - 1}}}{{{g^2}}}$, then we have
$$ \begin{align} 2{f^{'}}\left( m \right) &= \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\frac{{{a_i}{a_j}y_i^my_j^{m - 1}}}{{{g^2}}}\left[ {{x_j}\left( {\ln {x_j} - \ln {x_i}} \right)} \right]} } + \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\frac{{{a_i}{a_j}y_i^my_j^{m - 1}}}{{{g^2}}}\left[ {{x_i}\left( {\ln {x_i} - \ln {x_j}} \right)} \right]} } \\ &\ge \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\varepsilon \left[ {{x_j}\left( {\ln {x_j} - \ln {x_i}} \right) + {x_i}\left( {\ln {x_i} - \ln {x_j}} \right)} \right]} } \\ &\ge \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\varepsilon \left[ {\left( {{x_i} - {x_j}} \right)\left( {\ln {x_i} - \ln {x_j}} \right)} \right]} } \end{align} $$
Because $\forall\, u, v \in \mathbb{N}$, ${x_u}\left( {\ln {x_u} - \ln {x_v}} \right) + {x_v}\left( {\ln {x_v} - \ln {x_u}} \right) = \left( {{x_u} - {x_v}} \right)\left( {\ln {x_u} - \ln {x_v}} \right) > 0 $, thus $f^{'}({m})>0$. This completes the proof for the first part.
There might be purely algebraic proofs (maybe related to the Rearrangement Inequality), but the above proof works for now.
Actually, using the second part, we can easily derive the second part. Since $f$ is monotonically increasing with $m$, the maximum of $f$ is achieved when $m \rightarrow \infty$ where $w_k=1$ and $w_i=0,\; \forall \, i \ne k$.