A one-sided bound for √(1 + x) from the Lagrange remainder

Fourth derivative. For $f(x)=\sqrt{1+x}$, $f^{(4)}(x)=-\frac{15}{16}(1+x)^{-7/2}<0$.

Taylor–Lagrange. There is $c\in(0,x)$ with
$$\sqrt{1+x}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}+\frac{f^{(4)}(c)}{24}x^4,$$
and the last term is $\le0$. This is the inequality.

Enclosure. Moreover $|f^{(4)}(c)|\le\frac{15}{16}$, so the error is at most $\frac{15}{16\cdot24}x^4=\frac{5}{128}x^4$. For $x=0.2$: $\frac5{128}\cdot0.0016=6.25\cdot10^{-5}$. As $P_3(0.2)=1.0955$,
$$1.09543\le\sqrt{1.2}\le1.0955.$$
(True value: $1.095445\dots$)

Remark. The sign of the Lagrange remainder gives a *one-sided* inequality for free: this is how polynomial inequalities such as $\cos x\ge1-\frac{x^2}2$ are proved.