<s>[INST] 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) ?
 [/INST] From (A), for $x>0$ we have: 
$0<\sqrt{x+1}-1<\frac{1}{2}x\Rightarrow {(\sqrt{x+1}-1)}^2<\frac{1}{4}x^2\Rightarrow {(\sqrt{x+1})}^2+1^2-2\sqrt{x+1}<\frac{1}{4}x^2 \Rightarrow \sqrt{x+1}>1+\frac{x}{2}-\frac{x^2}{8}$
</s>