<s>[INST] Proving a complicated looking inequality in a simple way This is again a search for alternative proofs:
Let $0 <s \le 1$, and suppose that $0 <a,b $ satisfy 
$$ ab=s,a+b=1+\sqrt{s}. \tag{1}$$
I have a proof for the assertion
$$ 2(1-\sqrt s)^3 \le |a-1|^3+|b-1|^3, \, \, \, \text{for every } \, s \ge \frac{1}{9}$$
but it is rather involved. 
Actually, I am sure that the lower bound of $ \frac{1}{9}$, is not tight; the inequality holds for some $s > s^*$ where $s^* < \frac{1}{9}$.
Define $F(x,y):=|x-1|^3+|y-1|^3$. My proof is based on finding the global minimum $ \min_{xy=s} F(x,y)$.
However, here we need to show "only that" $F(\sqrt s, \sqrt s) \le F(a,b)$ for the specific $a,b$ described above in $(1)$.

Is there a way to prove this inequality "directly", without solving the harder global optimization problem?

Bonus: Is there a natural way to find the exact threshold $s^*$?
Edit:
There are now some very nice answers. I still wonder whether one can prove this without solving explicitly the quadratic described implicitly in $(1)$.
Here is an elementary proof for when $s \ge \frac{4}{9}$:
Suppose that $a \ge b$. The conditions on $a,b$ easily imply that $a \ge 1$, hence $s=ab \ge b$. Thus, we have
$$ b \le s \le \sqrt s \le 1 \le a.$$ 
So, replacing $b$ with $\sqrt s$ clearly lowers the value of $F$, since we get closer to $1$. Now it is beneficial to replace $a$ by $\sqrt s$ when
$$|\sqrt s -1|=1-\sqrt s \le a-1 \iff 2-\sqrt s \le a \iff  4-2\sqrt s \le 2a. \tag{2}$$
Solving explicitly the quadratic
$ a^2-(1+\sqrt s)a+s=0$, we get (assuming $a \ge b$) that
$$ a=\frac{1}{2}(1+\sqrt s+\sqrt{1+2\sqrt s-3s}).$$ Thus, inequality $(2)$ beceoms
$$
4-2\sqrt s \le 1+\sqrt s+\sqrt{1+2\sqrt s-3s},
$$
or $3-3\sqrt s \le \sqrt{1+2\sqrt s-3s}$. Squaring this and simplifying gives
$$
3s-5\sqrt s +2 \le 0,
$$
which holds exactly for $\frac{4}{9} \le s \le 1$.
 [/INST] Let $\sqrt{s}=t$. 
Thus, $\frac{1}{3}\leq t\leq 1.$
Since $a$ and $b$ are roots of the equation $$x^2-(1+t)x+t^2=0,$$ we need to prove that
$$\left|\frac{1+t+\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3+\left|\frac{1+t-\sqrt{(1+t)^2-4t^2}}{2}-1\right|^3\ge2(1-t)^3$$ or
$$\left|\frac{\sqrt{(1-t)(1+3t)}-(1-t)}{2}\right|^3+\left|\frac{\sqrt{(1-t)(1+3t)}+(1-t)}{2}\right|^3\ge2(1-t)^3$$ or
$$\left(\sqrt{1+3t}-\sqrt{1-t}\right)^3+\left(\sqrt{1+3t}+\sqrt{1-t}\right)^3\ge16\sqrt{(1-t)^3}$$ or 
$$\sqrt{(1+3t)^3}+3(1-t)\sqrt{1+3t}\ge8\sqrt{(1-t)^3}$$ or
$$(1+3t)^3+6(1-t)(1+3t)^2+9(1-t)^2(1+3t)\ge64(1-t)^3$$ or
$$4t^3-12t^2+15t-3\ge0,$$ which is true even for $t\ge\frac{1}{4}:$
$$4t^3-12t^2+15t-3=4t^3-t^2-11t^2+\frac{11}{4}t+\frac{49}{4}t-\frac{49}{16}+\frac{1}{16}=$$
$$=(4t-1)\left(t^2-\frac{11}{4}t+\frac{49}{16}\right)+\frac{1}{16}>0.$$
</s>