This starts with the observation in my comment above, quoted here:
"Since $\sum_i a_i=\sum_i b_i=1$, we can rewrite the inequality as $\sum_i a_i / ∑_{j\neq i}a_j \leq \sum_i b_i / \sum_{j \neq i} b_j=\sum_i a^2 / \sum_{j \neq i}a^2$. Then since both sides of this are constant under the transformation $a_i, \cdots, a_n \to \lambda a_i, \cdots, \lambda a_n$, it suffices to solve this new inequality without the condition $\sum_i a_i = 1$."
Therefore this problem is equivalent to showing that $f(1) \leq f(2)$, where
$$f(x) = \sum_i \frac{a_i^x}{\sum_{j \neq i}a_j^x}$$
and where we assume that $a_i > 0 \,\forall\, i$ (but not necessarily that $\sum_i a_i = 1$). It is sufficient to show that $f(x)$ is non-decreasing for $x > 0$. We calculate
$$f'(x) = \sum_i \frac{a_i^x \sum_{j \neq i} a_j^x \log \frac{a_i}{a_j}}{( \sum_{j \neq i} a_j^x )^2}$$
Letting $c_{i,x} = \sum_{j \neq i} a_j^x$, we can write this as:
$$f'(x) = \sum_{i \neq j} \frac{ (a_i a_j)^x \log \frac{a_i}{a_j}}{c_{i,x}^2} $$
$$= \sum_{i < j} (a_i a_j)^x \left(\frac{1}{c_{i,x}^2} \log\frac{a_i}{a_j} + \frac{1}{c_{j,x}^2} \log\frac{a_j}{a_i} \right)$$
$$= \sum_{i < j} (a_i a_j)^x \log\frac{a_i}{a_j} \left(\frac{1}{c_{i,x}^2} - \frac{1}{c_{j,x}^2} \right)$$
Assuming without loss of generality that $a_i \leq a_j$, it follows that $c_{i,x} \geq c_{j,x}$, so $\log\frac{a_i}{a_j} \leq 0$, $\frac{1}{c_{i,x}^2} - \frac{1}{c_{j,x}^2} \leq 0$, and thus $\log\frac{a_i}{a_j} \left(\frac{1}{c_{i,x}^2} - \frac{1}{c_{j,x}^2} \right)$ is non-negative. We also know that $(a_i a_j)^x \geq 0$, so therefore each term in the above summation is non-negative, and thus $f'(x) \geq 0$. QED