Deducing an inequality from a Taylor series expansion Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$
by noting that all the non linear terms are negative. Can the left side inequality be read-off the expansion by a similar reasoning? Please do not try to prove the left side inequality by other means (such as minimizing $\sqrt{1-x} - 1 + \frac{x}{2} + \frac{x^2}{2}$ using derivatives).

We need to prove that
$$(2-x-x^2)^2\leq4(1-x),$$ which is
$$x^2(3+x)(1-x)\geq0.$$
