Let $0 <s \le 1$, and suppose that $0 <b \le a$ satisfy $$ ab=s,a+b=1+\sqrt{s}.$$
Then $a \ge 1$.
I have a proof for this claim (see below), but I wonder if there are easier or alternative proofs.
In particular, my proof is based on explicit computation of $a,b$ in terms of $s$ (solving the quadratic). Can we avoid that?
The proof:
We have $a+\frac{s}{a}=1+\sqrt s$, or
$$ a^2-(1+\sqrt s)a+s=0,$$
which implies (since we assumed $a \ge b$) that $$ a=\frac{1}{2}(1+\sqrt s+\sqrt{1+2\sqrt s-3s}).$$
Thus, $a \ge 1$ iff $$\sqrt s+\sqrt{1+2\sqrt s-3s} \ge 1 \iff \\ 1+2\sqrt s-2s+2\sqrt s \sqrt{1+2\sqrt s-3s} \ge 1 \iff \\ \sqrt s-s+\sqrt s \sqrt{1+2\sqrt s-3s} \ge 0 \iff \\ 1-\sqrt s+ \sqrt{1+2\sqrt s-3s} \ge 0 \iff \\ \sqrt{1+2\sqrt s-3s} \ge \sqrt s-1.$$
(We passed from the first line to the second line by squaring).
The last inequality clearly holds, since the LHS $\ge 0$, and the RHS is $\le 0$. (since we assumed $s \le 1$).
Hint: $(a-1)(b-1) = ab - a - b + 1 = s - \sqrt{s} \leq 0 $.
Hence, conclude that $ a \geq 1 \geq b $, with equality when $ s = 1$.