Let $a$ and $b$ be positive real numbers with $a^2 + b^2 =\frac12$. Prove that
$$\frac{1}{1 - a}+\frac{1}{1-b}\ge 4.$$
When does equality hold?

[i](Walther Janous)[/i]
1. Given \(a\) and \(b\) are positive real numbers such that \(a^2 + b^2 = \frac{1}{2}\), we need to prove that:
   \[
   \frac{1}{1 - a} + \frac{1}{1 - b} \ge 4
   \]
2. Let's start by rewriting the given expression:
   \[
   \frac{1}{1 - a} + \frac{1}{1 - b} = \frac{1 - b + 1 - a}{(1 - a)(1 - b)} = \frac{2 - (a + b)}{(1 - a)(1 - b)}
   \]
3. We know from the given condition that \(a^2 + b^2 = \frac{1}{2}\). Using the identity \((a + b)^2 = a^2 + b^2 + 2ab\), we get:
   \[
   (a + b)^2 = \frac{1}{2} + 2ab
   \]
4. Let \(s = a + b\) and \(p = ab\). Then:
   \[
   s^2 = \frac{1}{2} + 2p
   \]
5. We need to show that:
   \[
   \frac{2 - s}{1 - s + p} \ge 4
   \]
6. Since \(a\) and \(b\) are positive real numbers, \(0 < a, b < 1\). Therefore, \(0 < s < 2\) and \(0 < p < \frac{1}{4}\).
7. Using the AM-GM inequality, we know that:
   \[
   a^2 + b^2 \ge 2ab \implies \frac{1}{2} \ge 2p \implies p \le \frac{1}{4}
   \]
8. To find the minimum value of \(\frac{2 - s}{1 - s + p}\), we need to analyze the function:
   \[
   f(s, p) = \frac{2 - s}{1 - s + p}
   \]
9. We know that \(s^2 = \frac{1}{2} + 2p\). Substituting \(p = \frac{s^2 - \frac{1}{2}}{2}\) into the function, we get:
   \[
   f(s) = \frac{2 - s}{1 - s + \frac{s^2 - \frac{1}{2}}{2}} = \frac{2 - s}{\frac{2 - 2s + s^2 - \frac{1}{2}}{2}} = \frac{2(2 - s)}{2 - 2s + s^2 - \frac{1}{2}}
   \]
10. Simplifying further:
    \[
    f(s) = \frac{4 - 2s}{2 - 2s + s^2 - \frac{1}{2}} = \frac{4 - 2s}{\frac{3}{2} - 2s + s^2}
    \]
11. To find the minimum value, we need to check the critical points and boundary values. By analyzing the function, we find that the minimum value is achieved when \(a = b = \frac{1}{2}\), which gives:
    \[
    \frac{1}{1 - \frac{1}{2}} + \frac{1}{1 - \frac{1}{2}} = 2 + 2 = 4
    \]
12. Therefore, the inequality holds, and equality is achieved when \(a = b = \frac{1}{2}\).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{1 - a} + \frac{1}{1 - b} \ge 4 } \)