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Czech-Slovak-Polish Match

Algebra Difficulty 6.0 AIME, harder Prove it Czech-Polish-Slovak Mathematical Match

Determine all triples (a,b,c)(a, b, c) of positive real numbers satisfying the following system of equations
ab−c=a,bc−a=b,ca−b=c. \begin{aligned} a\sqrt{b} - c &= a, \\ b\sqrt{c} - a &= b, \\ c\sqrt{a} - b &= c. \end{aligned}

Solution

Without loss of generality we can suppose a=max⁡{a,b,c}a = \max\{a, b, c\}. From the first equation of the given system we obtain
c(b−1)≤a(b−1)=c,i.e.b≤4. c(\sqrt{b} - 1) \le a(\sqrt{b} - 1) = c, \quad \text{i.e.} \quad b \le 4.
Similarly, from the second equation of the given system we have
b(c−1)=a≥b,thusc≥4. b(\sqrt{c} - 1) = a \ge b, \quad \text{thus} \quad c \ge 4.
Using these results together with the third equation we get
4≤c≤c(c−1)≤c(a−1)=b≤4 4 \le c \le c(\sqrt{c} - 1) \le c(\sqrt{a} - 1) = b \le 4
which yields a=b=c=4a = b = c = 4.

Conclusion. The unique solution of the given system of equations is therefore (a,b,c)=(4,4,4)(a, b, c) = (4, 4, 4).

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