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If only PA =1/ MA, then μB = νB =0 and Eqs. 20 21 22 23 hold like in the previous cases. The analog of Eq. 16 for i=A now imposes that MA =1,2 in order to have non-negativity of the Lagrange multipliers. The solution for MA =1 gives Eq. 30, which is thus proven to be valid in principle for all values of c, while the solution for MA =2 can only hold when c≤1/2. If Pi =1/ Mi for both A and B, a similar reasoning yields Mi =1,2 for i=A,B, and the resulting bounds are clearly nonoptimal.
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If only PA =1/ MA, then μB = νB =0 and Eqs. 20 21 22 23 hold like in the previous cases. The analog of Eq. 16 for i=A now imposes that MA =1,2 in order to have non-negativity of the Lagrange multipliers. The solution for MA =1 gives Eq. 30, which is thus proven to be valid in principle for all values of c, while the solution for MA =2 can only hold when c≤1/2. If Pi =1/ Mi for both A and B, a similar reasoning yields Mi =1,2 for i=A,B, and the resulting bounds are clearly nonoptimal.
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