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# Understand the problem

Find, with justification, all positive real numbers   $a,b,c$   satisfying the system of equations:    $$a\sqrt{b}=a+c,b\sqrt{c}=b+a,c\sqrt{a}=c+b.$$
##### Source of the problem
SMO (senior)-2014 stage 2 problem 2

Number Theory
Easy
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

Given all three relations are cyclic and symmetric . So without loss of generality it can be assumed that $a \geq b \geq c >0$ .     Then proceed .               [ Note $(0, 0, 0)$ can’t be a solution since $a , b , c$ are positive reals .]
So $a \sqrt b = a + c \Rightarrow a(\sqrt b – 1) = c \ [and \ we \ have \ a \geq c] \Rightarrow ( \sqrt b – 1 ) \leq 1 \Rightarrow b \leq 4$
Similarly $b \sqrt c = b + a \Rightarrow b(\sqrt c – 1) = a \ [and \ we \ have \ a \geq b ] \Rightarrow \sqrt c – 1 \geq 1 \Rightarrow c \geq 4$

Till now we have $b \leq 4 \ and \ c \geq 4$ , but we assumed that $b \geq c$ . So it is clear that $b = c =4$  $\Rightarrow a = 4$ also. So the only triplet $(a , b , c)$ is $(4,4 ,4)$ .

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