How to prove $1+{x\over 2}-{x^2 \over 8}<\sqrt{1+x}$ for all $x>0$? How to prove $1+(1/2)x-(1/8)x^2<\sqrt{1+x}$ for $x>0$ by Taylor Expansions up to $n$ order or Mean Value Theorem? I tried to apply MVT on $\sqrt{1+x}$ and get $\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x$ for $\xi\in (0,x)$. How to do next?

Taylor's formula isn't really required:
Set $\;f(x)=1+\dfrac x2-\dfrac{x^2}8$, $\;g(x)=\sqrt{1+x}$. We have
\begin{align}
f'(x)&=\frac12-\frac x4,&g'(x)&=\frac1{2\sqrt{1+x}},\\
f''(x)&=-\frac 14,&g''(x)&=-\frac1{4(1+x)^{3/2}}.
\end{align}
Observe that $\;f(0)=g(0)=1$, $\;f'(0)=g'(0)=\dfrac12$, $\;f''(0)=g''(0)=-\dfrac14$, and that $f''(x)<g''(x)< 0$ for all $x>0$.
A well-known corollary of the mean Value theorem ensures first that $f'(x)<g'(x)$, and thence that $f(x)<g(x)$ for all $x>0$.
