Suppose I have $w_i>0$ and $x_i>0$ for $i=1,2,\ldots, n$. I know that,
$$\sum_{i} w_i < \sum_{i} x_i \leq 1. \tag{1} \label{eq1}$$
Can I say that,
$$\sum_{i} w_i w_i < \sum_{i} w_i x_i. \tag{2} \label{eq2}$$
Edit: I change the question as follows:
Can we say that, $\sum_{i} w_i \log\left(w_i\right) <\sum_{i} w_i \log\left(x_i\right)$?
Clearly, no. For example:
$x_{1}=1/100,x_{2}=98/100, w_{1}=98/100, w_{2}=1/1000000$
Clearly, in this scenario, $$\sum_{i}w_{i}^{2}>\sum_{i}w_{i}x_{i}.$$
Adding some more conditions may help to make the proposition correct.
If assuming $w$ and $x$ are both ascending sort ($w_{1}<w_2<w_3<\cdots<w_{n}$, and vice versa), this is still wrong. For instance, $x_{1}=49/100,x_{2}=50/100, w_{1}=1/10000000,w_{2}=98/100$.
Clearly, if $x_{i+1}-x_i>w_{i+1}-w_i$, this proposition will be justified. But I failed to see any other solutions.
It turns out that, given $\bar{x} = \sum_{i} x_i > \sum_{i} w_i = \bar{w}$ where $x_i>0$, $w_i>0$ for $i=1,2...,n$, then,
$$\bar{x} = \sum_{i} x_i > \sum_{i} w_i = \bar{w}$$ $$\iff \sum_{i} w_i \log x_i > \sum_{i} w_i \log w_i$$
Proof:
$$\sum_{i} x_i > \sum_{i} w_i$$
$$\iff \sum_{i} w_i(\frac{x_i}{w_i} - 1) > 0$$
Using the fact that, $x < 1- \log x, \forall x \in (0,1)$, then, $1 - \log \frac{x_i}{w_i} > \frac{x_i}{w_i}$. Substituting the upper bound for $\frac{x_i}{w_i}$ will only increase the functions value. Thus,
$$\iff \sum_{i} w_i \big(1 - \log \frac{x_i}{w_i} - 1 \big) > 0$$
$$\iff \sum_{i} w_i(\log \frac{x_i}{w_i}) > 0$$
$$\iff \sum_{i} w_i \log x_i > \sum_{i} w_i \log w_i) \tag*{$\blacksquare$} $$
