To solve the problem of finding the maximum value of (P = \frac{ab}{1 - c} + \frac{bc}{1 - a} + \frac{ca}{1 - b}) where (a, b, c > 0) and (a + b + c = 1), follow these steps:
Since (a + b + c = 1), we have (1 - c = a + b), (1 - a = b + c), (1 - b = c + a). Thus: [P = \frac{ab}{a + b} + \frac{bc}{b + c} + \frac{ca}{c + a}]
For any positive (x, y), (\frac{xy}{x + y} \leq \frac{x + y}{4}) (since (xy \leq \frac{(x + y)^2}{4}), dividing both sides by (x + y) gives this).
[ \sum{cyc} \frac{ab}{a + b} \leq \sum{cyc} \frac{a + b}{4} ] [ = \frac{(a + b) + (b + c) + (c + a)}{4} ] [ = \frac{2(a + b + c)}{4} = \frac{2 \times 1}{4} = \frac{1}{2}]
Equality holds when (a = b = c) (since (\frac{xy}{x + y} = \frac{x + y}{4}) iff (x = y)). Thus (a = b = c = \frac{1}{3}).
Answer: (\boxed{\dfrac{1}{2}})
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