Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$, prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$ Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$, 
prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$
Im thinking along the lines of binominal expansion as $\sqrt{1+x} = 1 + 0.5x - 0.125x^2 + ...$
But how im not sure how to continue
 A: You do not have a calculus tag, so the first solution below may be inappropriate. 
First Solution: Let 
$$f(x)=\sqrt{1+x}-(1+0.5x-0.125x^2).$$
We want to show that $f(x)\gt 0$ if $x\gt 0$. Note that $f(0)=0$. We will show that $f(x)$ is increasing.
We have
$$f'(x)=\frac{1}{2\sqrt{1+x}}-(0.5-0.25 x).$$
We will show that $f'(x)\gt 0$ if $x\gt 0$.
Since $\sqrt{1+x}\lt 1+0.5x$, we have 
$\frac{1}{2\sqrt{1+x}}\gt \frac{1}{2+x}$. We will show that
$$\frac{1}{2+x}-(0.5-0.25x)\gt 0.$$
Bring to  common denominator. We need to show that
$$1-(2+x)(0.5-0.25x)\gt 0.$$
This is clear, for $1-(2+x)(0.5-0.25x) =0.25x^2$.
Second Solution: If $1+0.5x-0.125x^2$ is negative the result is obvious. 
Note that if $x\ge 8$, then $1+0.5x-0.125x^2$ is negative. For then $1+0.5x-0.125x^2\le 1+0.5x-x\le -3$.
So we may assume that $1+0.5x-0.125x^2$ is positive, and also that $x\lt 8$.   It is then enough to show that $(1+0.5x-0.125x^2)^2\lt 1+x$. Do the squaring. We get
$$(1+x/2-x^2/8)^2=1+x+x^2/4-2(x^2/8)-2(x^3/16)+x^4/64.$$
We want to show this is $\lt 1+x$, so we need to show that
$$x^4/64-x^3/8\lt 0.$$
This is clearly true if $x\lt 8$.  
