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?

Let $f(x)=\sqrt {1+x}\;.$ For $x>0$ we have $$f(x)=f(0)+xf'(0)+x^2f''(y)/2$$ where $0<y<x.$ We have $f(0)=1$ and $f'(0)=1/2.$ We have $$f''(y)=-\frac {1}{4(1+y)^{3/2}}>-\frac {1}{4}  .$$ So  $\sqrt {1+x}=f(x)=1+x/2-x^2f''(y)/2>1+x/2-x^2/8.$
