Use Mean value theorem to prove the following inequality A) Use the Mean value theorem to prove that
\begin{equation}
\sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0
\end{equation}
B) Use result in A) to prove that 
\begin{equation}
\sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0
\end{equation}
Can someone give an answer for part B) ?

OK, I'll show how I'd solve a:
We have 
$$
f(x) = 1+ \frac{x}{2} - \sqrt{1+x} 
$$
Clearly $f(0) = 0$. Let's look at the derivative of $f$. If we show that the derivative is strictly positive for $x>0$, the function is strictly increasing, hence positive (in combination with $f(0)$) for $x>0$. We have 
$$
f'(x) = \frac{1}{2} - \frac{1}{2\sqrt{1+x}} = \frac{1}{2}\bigg(\frac{x}{\sqrt{1+x}} \bigg)
$$
Since $x>0$, the term in the brackets is always positive, hence the derivative is strictly positive and the function is increasing. Hence, 
$$
f(x)>0 \to 1+\frac{x}{2} > \sqrt{1+x}
$$ 
