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 $\frac{4}{9} \le s \le 1$.
 A: 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 $t\ge\frac{1}{4}:$
$$4t^3-12t^2+15t-3=4t^3-t^2-11t^2+\frac{11}{4}t+\frac{49}{4}t-\frac{49}{16}+\frac{1}{16}=$$
$$=(4t-1)\left(t^2-\frac{11}{4}t+\frac{49}{16}\right)+\frac{1}{16}>0.$$
A: By symmetry, we may take $a\leqslant b$ without loss of generality. Since the upper bound on $s$ ensures that $a$ and $b$ cannot both exceed $1$, we have $a\leqslant1$. It is convenient to transform the variables as follows: $$t:=1-\surd s,\qquad u:=1-a,\qquad v:=b-1.$$ Then the relationships between $a$, $b$, and $s$ become $$u-v=t,\qquad uv=t-t^2.$$Clearly $u\geqslant v$ and $u\geqslant0$. Also $v\geqslant0$ since $uv=t-t^2\geqslant0$ for $0\leqslant t<1$. Hence $u$ and $-v$ are the roots in $x$ of the quadratic equation $$x^2-tx+t^2-t=0,$$ where
$$u=\tfrac12\surd(4t-3t^2)+\tfrac12t,\qquad v=\tfrac12\surd(4t-3t^2)-\tfrac12t.$$
Let $$f(t):=u^3+v^3-2t^3.$$ Our task is to find the range of $t$ for which $f(t)\geqslant0$. Since $u^3+v^3=(u+v)[(u+v)^2-3uv]$, we have $$f(t)=t\surd(4t-3t^2)-2t^3.$$ In the given range $0\leqslant t<1$, this function initially increases from zero, attains its maximum $\frac14(\surd5-1)$ at $t=\frac12$, and then decreases to zero at $t=\alpha$, where $\alpha$ is the real root of $$t^3=1-\tfrac34t.$$ The corresponding value of $s$ is $(1-\alpha)^2,$ or approximately $0\!\cdot\!059354279$ according to my calculator.
A: The primary objective of this text is to give the following insight into this interesting issue : it boils down to the fact (see figure below) that a certain curve is above a certain straight line, boundary of the domain defined by homogenous inequation (4). 
I will make the following change of variables similar to what @John Bentin has done :
$$x:=a-1, \ \ \ y:=b-1, \ \ \ t:=\sqrt{s} \ \text{with} \ 0<t<1, \ \tag{1}$$
transforming the initial constraints into 
$$\begin{cases}x+y&=&t-1\\ \ \ \ xy&=&t(t-1)\end{cases}\tag{2}$$
As $x$ and $y$ play a symmetrical rôle, we may assume $x<y$.
Using (2), a rapid calculation shows that (solving quadratic equation $X^2-(t-1)X+t(t-1)=0$) :
$$\begin{cases}x&=&\frac12(t-1-\sqrt{\Delta})\\y&=&\frac12(t-1+\sqrt{\Delta})\end{cases} \ \ \ \text{with} \ \Delta=(3t+1)(1-t)>0\tag{3}$$
It is immediate to see that $x<0$ whereas $y>0$ due to the second relationship in (2).
Therefore, the inequality we have to establish 
$$2(1-t)^3 \le |x|^3+|y|^3$$
can be written
$$-2(x+y)^3 \leq -x^3+y^3\tag{4}$$
Consider now the plane with coordinates $(x,y)$. Let us plot in it, 


*

*the curve $(C_1)$ (in red) with parametric equations (3).

*the (frontier) curve $(C_2)$ (in blue) with implicit equation (4) in which the $\leq$ sign has been replaced by the $=$ sign :
$$-2(x+y)^3 = -x^3+y^3\tag{5}$$

Fig. 1 : Representation of curves defined by (3) and  (5). Please note that only the left hand side plane $x<0$ makes sense here.
Curve $(C_2)$ is a line. Not so surprizing in fact (see remark 1 below). Indeed, plugging $y=ux$ into (4) gives the following constraint on $u$ :
$$-2((1+u)x)^3 = (u^3-1)x^3 \ \ \iff \ \ -2(1+u)^3=u^3-1,\tag{6}$$
a third degree equation whose unique real root is $u_0 \approx -0.20406$, meaning that the equation of the line is approximately $y=-0.2x$.
Now, that we have well understood the nature of the frontier, we can infer that the region defined by inequation (4) is the half-plane situated above  the straight line we have found (one reason among others : point $(x,y)=(0,1)$ belongs to this region). 
It remains to prove, as suggested by the figure, that the red curve is entirely situated into this favorable region.
Remarks: 
1) The fact that the curve associated with (5) is a straight line can be explained differently by considering that it's homogeneous (if $(x,y)$ is on the curve, $(\lambda x, \lambda y)$ belongs as well to the curve) ; technically speaking, we could have as well divide its LHS and RHS by $x^3$, generating a 3rd degree equation with variable $u:=\tfrac{y}{x}$. 
2) (on an experimental basis) one can take $t \ge 0.2436...$ instead of $t \ge 1/3$.
