, 2014

Without loss of generality, we assume that $a_1 \ge a_2 \ge \dots \ge a_n$, so also $b_1 \ge b_2 \ge \dots \ge b_n$ and $0 < a_i, b_i < 1$. Let $S = \sum_{i=1}^n a_i^2$, then $b_i = a_i^2/S$. Note that $S = S \cdot \sum_{i=1}^n a_i$ and $\sum_{i=1}^n b_i = 1$. For $1 \le k \le n$ let
$$D_k = \sum_{i=1}^{k} (b_i - a_i),$$
in particular $D_n = 0$, because $\sum_{i=1}^k a_i = \sum_{i=1}^k b_i = 1$. We have for $1 \le k < n$
$$S \cdot D_k = \sum_{i=1}^{k} \sum_{j=1}^{n} a_i^2 a_j - \sum_{i=1}^{k} \sum_{j=1}^{n} a_i a_j^2 = \sum_{\substack{1 \le i \le k \\ k < j \le n}} a_i a_j (a_i - a_j) \ge 0$$
because all terms in the sum are non-negative. As $S > 0$, this implies that $D_k \ge 0$ for all $1 \le k \le n$. Note also that one of the terms in the sum, namely $a_1 a_n (a_1 - a_n)$, is positive unless $a_1 = a_2 = \dots = a_n$. So we have $D_k = 0$ for $1 \le k < n$ if $a_1 = a_2 = \dots = a_n$, and $D_k > 0$ for $1 \le k < n$ otherwise. Next observe that, for real numbers $a \ne 1$ and $b \ne 1$, we have
$$\frac{b}{1-b} - \frac{a}{1-a} = \frac{1}{(1-b)(1-a)} \cdot (b-a).$$
Setting $c_i = 1/(1-b_i)(1-a_i)$ this implies
$$D := \sum_{i=1}^{n} \frac{b_i}{1-b_i} - \sum_{i=1}^{n} \frac{a_i}{1-a_i} = \sum_{i=1}^{n} c_i (b_i - a_i) = \sum_{i=1}^{n-1} (c_i - c_{i+1}) D_i,$$
where we have used that $D_n = 0$. From $a_i \ge a_{i+1}$ and $b_i \ge b_{i+1}$ we get $c_i \ge c_{i+1}$, hence all terms in this sum are non-negative. This shows that $D \ge 0$ which is equivalent to the inequality we want to prove.
As for equality, we have $D = 0$ if and only if every term in the last sum equals 0. If $a_1 = a_2 = \dots = a_n$, then, as we have seen above, each $D_i = 0$ and so the sum is zero. If however not all numbers $a_i$ are equal, then we have seen that $D_i > 0$ and $c_i - c_{i+1} \ge 0$ for $1 \le i < n$. Therefore, all we need to show is that $c_i > c_{i+1}$ for one such $i$. If not all $a_i$ are equal to each other, there exists $1 \le i < n$ such that $a_i > a_{i+1}$ and $b_i \ge b_{i+1}$. For this $i$, the definition of $c_i$ indeed implies that $c_i > c_{i+1}$.
Thus the condition for equality is that all the numbers $a_i$ are equal to $1/n$.

If $p \ge 1$ is a real number, the function $F(x) = x^p$ is strictly convex on $[0, \infty)$. Hence, if $x_i \ge 0$, $\alpha_i \ge 0$ and $\sum_i \alpha_i = 1$, then
$$\left(\sum_i \alpha_i x_i\right)^p \le \sum_i \alpha_i x_i^p.$$