Proving a complicated looking inequality in a simple way

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 $t\ge\frac{1}{4}:$ $$4t^3-12t^2+15t-3=4t^3-t^2-11t^2+\frac{11}{4}t+\frac{49}{4}t-\frac{49}{16}+\frac{1}{16}=$$ $$=(4t-1)\left(t^2-\frac{11}{4}t+\frac{49}{16}\right)+\frac{1}{16}>0.$$

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.