Define $$x_{k+1}(t)=\frac{3x^4_k(t)+6(1-t)x_k^2(t)-(1-t)^2}{8x_k^3(t)},$$ with $x_0(t)=1$. It is not difficult to see $x_k(t)$ converges to $\sqrt{1-t}$, whose (Maclaurin expansion) has negative coefficients unless the first one. Let
$$x_k(t)=\sum\limits_{i =0}^{\infty}c_{k,i}t^{i}, \mid t\mid<1$$
Is it true $c_{k,i}\le0$ for all $k\ge 1, i\ge 1$?
This is not the first time I am fighting with positivity (nonnegativity). But this problem looks not natural enough for a standard technique, and it seems to me that the resulting sequence $x_k(t)$ is always between $\sqrt{1-t}$ and $1$, in the sense that the expansions of $x_k(t)-\sqrt{1-t}$ and $1-x_k(t)$ have nonnegative coefficients only.
I can only give a partial solution. After the change $$ z_k(t)=1-\frac{x_k(t)}{\sqrt{1-t}} $$ the recursion for $x_k(t)$ translates into $$ z_{k+1}=\frac{z_k^3(4-3z_k)}{8(1-z_k)^3} \quad\text{for}\quad k=0,1,2,\dots $$ with the initial data $$ z_0=1-\frac1{\sqrt{1-t}}=-\frac12t+O(t^2). $$ The recursion implies $z_{k+1}=z_k^3/2+O(z_k^4)$, therefore by induction on $k$ we obtain $$ z_k=-\frac{t^{3^k}}{2^{(3^{k+1}-1)/2}}+\text{higher terms}. $$ This implies that $$ x_k(t)=\sqrt{1-t}-\sqrt{1-t}z_k(t) $$ agrees with $\sqrt{1-t}$ up to the term $t^{3^k}$. In particular, as the expansion of $\sqrt{1-t}$ involves negative coefficients besides the constant term, we conclude that $c_{k,i}<0$ for $i=1,2,\dots,3^k$ in the expansion $$ x_k(t)=1+\sum_{k=1}^\infty c_{k,i}t^i. $$
