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\begin{align*}a^a\cdot b^b\cdot c^c\cdot d^d&=\frac12\\a+b+c+d&=1\end{align*}

How can we find solutions for this system of equations given that $a, b, c, d > 0$ ?

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  • $\begingroup$ Must the values be positive? $\endgroup$ – Calvin Lin Sep 17 '13 at 21:56
  • $\begingroup$ There seems to be infinitely many solutions, with one of the values quite close to 1. I don't think you can find a simple description of the solution set. $\endgroup$ – Calvin Lin Sep 17 '13 at 22:11
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    $\begingroup$ If you try $a = 0.97, b =c=d= 0.01$, you get 0.845.... If you try $ a = 0.7, b=c=d=0.1$, you get 0.390.... There has to be a solution in the line between the two. Slight perturbations would also yield a solution. $\endgroup$ – Calvin Lin Sep 17 '13 at 22:38
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The system you described is a system of 2 equations with 4 unknowns. To obtain a unique solution you need 4 independent equations. Since you have two, it means you have two free variables. You need to assume values for those two free variables and solve for the other two using the given equations.

Since it is a condition that all numbers are greater than zero and their summation is 1. The values of the assumed free variables has to be between zero and one

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