Taylor Series type inequality

Question

I need to show that for $x>0$, $1+\frac{x}{2} \ge \sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $.

I used the geometric/arithmetic mean inequality to show that $1+\frac{x}{2} \ge \sqrt{1+x}$ is indeed true. My issue lies in the second part of this.

In trying to show that $\sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $, I determined that this is a partial taylor series expansion and I attempted to play around with both inequalities to show that this is true but to no avail.

I would appreciate any help in solving this. Ideally I would like to do this analytically but any solutions using more complicated theorems are welcome.

Answer 1

$$0\leq\left(1+\frac{x}{2}\right)-\sqrt{1+x}=\frac{\left(1+\frac{x}{2}\right)^2-(1+x)}{1+\frac{x}{2}+\sqrt{1+x}}\leq\frac{\frac{x^2}{4}}{2\sqrt{1+x}}\leq\frac{x^2}{8}.$$

Answer 2

By Taylor's Theorem, for each $x>0$, there is $\zeta\in (0,x)$ such that

$$ \sqrt{1+x} = 1+ \frac{1}{2}x-\frac{1}{8}x^2+ \frac{1}{16}(\zeta+1)^{-\frac{5}{2}}x^3.$$

It is clear that for $x>\zeta>0$: $$ \frac{1}{16}(\zeta+1)^{-\frac{5}{2}}x^3 >0$$