Problem 1241

The function $f(x) = \frac{1}{1-x}$, $0 \le x < 1$, is strictly convex on $[0, 1)$, and so, if $\alpha_i$, $i = 1, 2, \dots, n$ are non-negative numbers that sum to $1$, then
$$f\left(\sum_{i=1}^{n} \alpha_i a_i\right) \le \sum_{i=1}^{n} \alpha_i f(a_i).$$

Solution 1. In particular, with $\alpha_i = a_i$,
$$\frac{1}{1 - \sum_{i=1}^{n} a_i^2} = f\left(\sum_{i=1}^{n} a_i a_i\right) \le \sum_{i=1}^{n} a_i f(a_i) = \sum_{i=1}^{n} \frac{a_i}{1 - a_i}.$$
But, by the Cauchy-Schwarz inequality,
$$1 = \left(\sum_{i=1}^{n} a_i\right)^2 \le n \sum_{i=1}^{n} a_i^2,$$
with equality iff $a_i = 1/n$, $i = 1, 2, \dots, n$. In other words,
$$\frac{n}{n-1} \le \frac{1}{1 - \sum_{i=1}^{n} a_i^2} \le \sum_{i=1}^{n} \frac{a_i}{1 - a_i},$$
with equality iff $a_i = 1/n$, $i = 1, 2, \dots, n$.

Solution 2. Using $\alpha_i = 1/n$, convexity gives
$$\frac{n}{n-1} = f\left(\frac{1}{n}\right) = f\left(\frac{1}{n} \sum_{i=1}^{n} a_i\right) \le \frac{1}{n} \sum_{i=1}^{n} f(a_i),$$
with equality iff the $a_i$ are equal to each other. Hence
$$n + \frac{n}{n-1} = \frac{n^2}{n-1} \le \sum_{i=1}^{n} f(a_i) = \sum_{i=1}^{n} \left(1 + \frac{a_i}{1-a_i}\right) = n + \sum_{i=1}^{n} \frac{a_i}{1-a_i}$$
and this is equivalent to the desired inequality. Moreover, the inequality is strict unless the $a_i$ are equal to each other.