Saudi Arabia Mathematical Competitions 2012

The given relation is equivalent to $$(b+ac+a+bc)(a+b)=ab+c(a^2+b^2)+ab{c}^2,$$ so therefore $$(a+b{)}^{2}c+(a+b{)}^2=ab(c^2+1)+c(a^2+b^2).$$ It follows $$\begin{array}{rl}(a+b{)}^2& =ab(c^2+1)+c[a^2+b^2-(a+b{)}^2]\\ & =ab(c^2+1)-2abc=ab(c-1{)}^2.\end{array}(1)$$ If $c=1$, from the given relation we get $$\frac{1}{a+b}+\frac{1}{b+a}=\frac{1}{a+b},$$ which is not possible. Therefore, we have $c\ne 1$, and from (1) we obtain $$ab={\left(\frac{a+b}{c-1}\right)}^2.(2)$$ Using (2), it follows $$\begin{array}{rl}(c-3)(c+1)& =(c-1{)}^2-4=\frac{(a+b{)}^2}{ab}-4=\frac{(a-b{)}^2}{ab}\\ & ={\left[\frac{(a-b)(c-1)}{a+b}\right]}^2.\end{array}$$ Finally, $$\sqrt{\frac{c-3}{c+1}}=\frac{\sqrt{(c-3)(c+1)}}{|c+1|}=\frac{|a-b|\cdot |c-1|}{|c+1|\cdot |a+b|}\in \mathbb{Q}.$$

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Alternative solution.

Assume $b=0$. Then we get $$\frac{1}{a}+\frac{1}{ac}=\frac{1}{a},$$ which is not possible. It follows that $b\ne 0$, so we can divide by $b$ in the given relation and obtain $$\frac{1}{\frac{a}{b}+c}+\frac{1}{1+\frac{a}{b}c}=\frac{1}{\frac{a}{b}+1},\frac{a}{b}\in \mathbb{Q}.(1)$$ Let $\frac{a}{b}=r\in \mathbb{Q}$. Then (1) is equivalent to $$\frac{1}{r+c}+\frac{1}{1+rc}=\frac{1}{r+1}.(2)$$ From (2) we obtain $(1+r+rc+c)(r+1)=(r+c)(rc+1)$, hence $r^2-[(c-1{)}^2-2]r+1=0$. This is a quadratic equation with rational roots, hence the discriminant $\Delta$ must be a perfect square of a rational number. We have $$\begin{array}{rl}\Delta & =[(c-1{)}^2-2{]}^2-4=[(c-1{)}^2-4](c-1{)}^2\\ & =(c-3)(c+1)(c-1{)}^2=\frac{c-3}{c+1}(c+1{)}^2(c-1{)}^2,\end{array}$$ implying that $\frac{c-3}{c+1}$ is a perfect square of a rational number, and we are done.