Question: How can I use the mean value theorem to prove that the inequality
$$1 + \frac{x}{2\sqrt{1+x}} < \sqrt{1 + x} < 1 + \frac{x}{2}$$ 
holds for all $x\in (-1,0)$?

Answer: 
Hint:

Write the inequalities as
$$
\frac{1}{2\sqrt{1+x}}>\frac{\sqrt{1+x}-1}{x}>\frac{1}{2}.\quad \text{Warning $x<0$}
$$
Let $f(x)=\sqrt{1+x}$. Mean value theorem implies that $\frac{\sqrt{1+x}-1}{x}=f'(\theta)$ for $\theta\in (x,0)$. I will left the rest as your work.
Let 
\begin{align\*}
f(x)= \sqrt{1 + x} -1 - \frac{x}{2\sqrt{1+x}} \\
g(x)= 1 + \frac{x}{2} - \sqrt{1 + x}
\end{align\*}
We want to show $f(x)>0$ and $g(x)>0$ when $x \in (-1,0)$.

Let's focus on $g(x)$. $g(0)=0$ and 
$$
g'(x)=\frac{1}{2}-\frac{1}{2\sqrt{1+x}}< 0
$$
so $g(x)$ is decreasing from -1 to 0 so increasing from 0 to -1. Hence $0=g(0)<g(x)$ on $(-1,0)$. Similarly, $f(x)>0$.