Number theory problem, fractions and gcd The problem says:If a and b are positive integers...

If $gcd(a,b)=1$, then clearly $\sqrt{a+b}\ge gcd(a,b)$. 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, $a+b\ge d^2⟹\sqrt{a+b}\ge d$