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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

# Start with hints

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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#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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