Number theory problem, fractions and gcd The problem says:

If $a$ and $b$ are positive integers such that $\dfrac{a+1}{b}+\dfrac{b+1}{a}$ is an integer, then show that $\sqrt{a+b}\ge \gcd(a,b)$.

Adding $\frac{2ab}{ab}$ to $\frac{a+1}{b}+\frac{b+1}{a}$ yields that ab divides (a+b+1)(a+b), but I haven´t been able to continue from there.

If $\gcd(a,b)=1$, then clearly $\sqrt{a+b} \geq \gcd(a,b)$.  Now assume $\gcd(a,b)=d>1$.  Then if $\frac{a+1}{b}+\frac{b+1}{a}=n$, then by your simplification, we have $$(a+b+1)(a+b)=(n+2)ab$$Note that $d^2$ divides the righthand side, so it must also divide the lefthand side.  Since $d$ divides $a$ and $d$ divides $b$, $d$ must divide $a+b$.  Since $a+b+1$ and $a+b$ are coprime, and since $d$ divides $a+b$, we must also have $d^2$ dividing $a+b$.  In particular, $a+b\geq d^2 \implies \sqrt{a+b}\geq d$.  
