Given $d = \min(\gcd(a, b+1),\gcd(a+1,b))$, prove $d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$ The question is stated above:

Given $d = \min(\gcd(a, b+1),\gcd(a+1,b))$, prove $d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$.

Here's what I've done so far:
$$d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$$
then
$$2d + 1 \leqslant \sqrt{4a+4b+5}$$
Squaring both sides, we have
$$4d^2+4d+1 \leqslant 4a+4b+5$$
$$4d^2+4d \leqslant 4a+4b+4$$
$$d^2+d \leqslant a+b+1$$
From here I'm stuck and unable to continue. I've also noticed that the sum $a+b+1$ can have some relations with the two GCDs, since $a+b+1=(a+1)+b=a+(b+1)$, but I'm unable to work out how.

You are nearly there. Let $d<d_1$ be the two gcds. Then $a+b+1$ is divisible by $d$ as well as by $d_1$ and $gcd(d,d_1)=1$. So $a+b+1 \geq d d_1$.
