In an algorithm to derive the rate of convergence, I come across a technical problem. Numerical experiments indicate the following proposition, yet I don't have a proof. Can anyone give me some suggestion?

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 $x_0(t)=1$. Let

$$x_k(t)=\sum\limits_{i =0}^{\infty}c_{k,i}t^{i}, \mid t\mid<1$$

Then $c_{k,i}\le0$ for all $k\ge 1, i\ge 1$.

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 $k\ge 1, i\ge 1$?