[M.B.E.]

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The solution is : bet all your hands X > c with

c = ( b / (n*a + b) ) ^ (1/(n-1))

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Not quite. The correct answer is: bet all your hands X > c with

c = ( b/ ((n-1)*a + b) ) ^ (1/(n-1))

It looks nicer if you say m is the number of your opponents = n-1. Then

c = (b/(am+b))^(1/m).

The error in cbloom's proof is in the assumption that if you fold your EV is 0. You actually have positive EV when you fold, because if everyone else also folds it's a push and you get your ante back. The probability of that happening is c^m = P.

Then, using cbloom's reasoning:

P*(n*a) + (1-P)*(-b) = P*a
P*(n*a + b) - b = P*a
P*(n*a + b - a) - b = 0
P = b/(n*a + b - a)
P = b/((n-1)*a + b)= b/(am+b)

So if we take DS's 3-player game with a=1, b=2, m=2, then c is the square root of 2/(2+2), or sqrt(0.5) = 0.7071. Incidentally that's the same answer posted by emanon.

For the case of a=1, b=2, m=1 (heads up), we get:
c = 2/(1+2) = 0.66667.