Show that the following inequality holds when $x>0$ $\require{cancel}$

Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$

I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the inequality yields
$$1+\frac{x}{2}-\frac{x^2}{8}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<1+\frac{x}{2}$$
which is equivalent to
$$\begin{cases}
\cancel{1+\frac{x}{2}-\frac{x^2}{8}}\cancel{<1+\frac{x}{2}-\frac{x^2}{8}}+\frac{x^3}{16 (\xi+1)^{5/2}}\quad(1)\\
\cancel{1+\frac{x}{2}}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<\cancel{1+\frac{x}{2}}\qquad\,\,\,\,\,\, (2)
\end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$\frac{x^3}{16 (\xi+1)^{5/2}}<\frac{x^2}{8}$$
But $$\displaystyle{\max_{0\leq\xi\leq x}\Bigg\{\frac{1}{16 (\xi+1)^{5/2}}\Bigg\}}=\displaystyle{\min_{0\leq\xi\leq x}{\Big\{16 (\xi+1)^{5/2}}\Big\}}$$
which occurs at $\xi=0$, and thus, for $(2)$, it is left to prove that 
$$\frac{x^3}{16}<\frac{x^2}{8}$$
but this happens for $x<0\,\vee\,0<x<2$.
Instead, if I use $\xi=x$ then inequality $(2)$ becomes
$$\frac{x^3}{16 (x+1)^{5/2}}<\frac{x^2}{8}$$
which holds for $-1<x<0\,\vee\,x>0$, and thus coincides with the restriction given by the  problem.
Would it be correct to take $\xi=x$ rather than $0$? Is this approach correct at all?
 A: The right inequality:
We need to prove that:
$$1+x<\left(1+\frac{x}{2}\right)^2$$ or
$$1+x<1+x+\frac{x^2}{4}.$$
The left inequality.
Let $x+1=t^2,$ where $t>1$.
Thus, we need to prove that:
$$1+\frac{t^2-1}{2}-\frac{(t^2-1)^2}{8}<t$$ or
$$(t-1)^3(t+3)>0.$$
A: At $x=0$, all three members equal $1$. Then we can take the derivative and
$$\frac12-\frac x4<\frac1{2\sqrt{x+1}}<\frac12.$$
Again, we have equality at $x=0$ and
$$-\frac14<-\frac1{4(x+1)^{3/2}}<0.$$
This final bracketing is obvious.
A: For the left inequality:
$$
1 + \frac{x}{2} - \sqrt{1 + x} = \frac{\left(1 + \frac{x}{2}\right)^2 - (1 + x)}{1 + \frac{x}{2} + \sqrt{1 + x}} < \frac{\frac{x^2}{4}}{2} = \frac{x^2}{8}.
$$
A: Well the inequality can be proved using algebra by squaring it. A proof via Taylor's theorem is as follows. We have a number $c\in(0,x)$ such that $$\sqrt {1+x}=1+\frac{x}{2}-\frac{x^2}{8}(1+c)^{-3/2}$$ The desired inequality follows because $0<(1+c)^{-3/2}<1$.
Your approach is similar but unnecessarily uses third derivatives and complicates the situation. 
