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.

Let $d_1=\gcd(a,b+1)$ and $d_2=\gcd(a+1,b)$.
Then, as you've mentioned already, $a+b+1=a+(b+1)=(a+1)+b$, so $d_1$ and $d_2$ are divisors of the $a+b+1$. Since $d_1\mid a$ and $d_2\mid a+1$ we also have $\gcd(d_1,d_2)=1$. Thus, $d_1d_2\mid a+b+1$. In particular, $d_1d_2\le a+b+1$.
Finally, if $d_1=d_2$, then $d_1=d_2=1$ and the statement is trivial and otherwise due to $d=\min\{d_1,d_2\}$ we have
$$
a+b+1\ge d_1d_2\ge d(d+1),
$$
as desired.
