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?
 A: $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{\left(\sqrt{x+1}\right)'''_{x=\xi}x^3}{6}$$ and since $\left(\sqrt{x+1}\right)'''>0$ for all $x>0$, we are done!
A proof without Taylor series. 
If $1+\frac{x}{2}-\frac{x^2}{8}<0$ or $x>2+2\sqrt3$ then  it's obvious.
But for $0<x\leq2+2\sqrt3$ it's enough to prove that
$$(x^2-4x-8)^2<64(1+x)$$ or
$$x^3(x-8)<0.$$
Done!
A: 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  $\sqrt {1+x}=f(x)=1+x/2-x^2f''(y)/2>1+x/2-x^2/8.$
A: 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 $x>0$.
A: Here is another way to prove the inequality without using Taylor expansion. Let
$$ f(x)=\frac{1}{\sqrt{1+x}}+\frac12x. $$
Then
$$ f'(x)=-\frac{1}{2(\sqrt{1+x})^3}+\frac12>0, \text{ for } x>0 $$
namely $f(x)$ is increasing for $x>0$. So
$$ \frac{1}{\sqrt{1+x}}+\frac12x>1$$
or
$$ \frac{1}{\sqrt{1+x}}>1-\frac12x. $$
Integrating from $0$ to $x$ gives
$$ \int_0^x\frac{1}{\sqrt{1+t}}dt>\int_0^x(1-\frac12t)dt $$
from which one has
$$ \sqrt{1+x}>1+\frac12x-\frac18x^2. $$
A: As requested, with Taylor series. Consider $f(x)=\sqrt{1+x}$. It's well known that $f$ has a Taylor series centred at $0$ with radius of convergence $1$, which begins 
$$f(x) =\sum_{k=0}^{\infty}c_{k}x^{k} = 1+\frac{1}{2}x-\frac{1}{8}x^{2}+\ldots$$
You can get an exact expression for the coefficients (it's something like $c_{k} = \frac{(-1)^{k}}{k+1}{2k \choose k}$), but all that matters is that $\sum_{k=0}^{\infty}|c_{k}|$ converges, the coefficients are alternating in sign, and $|c_k x^k|$ forms a decreasing sequence. Taylors theorem tells us that for all $x \in (0,1)$, 
$$f(x) - \sum_{k=0}^{n}c_{k}x^{k}= c_{n+1}\xi^{n+1}$$
for some $\xi \in (0,1)$ depending on $x$. In any case, this difference is bounded above (taking the modulus) by $|c_{n+1}|$, independently of $\xi$. So for every $x\in(0,1)$, 
$$f(x) = \sum_{k=0}^{\infty}c_{k}x^{k}$$
You can also readily see that for $k>0$, $c_{k} = (-1)^{k+1}|c_{k}|$ - that is, the coefficients alternate. I claim that this implies that, for every $n>0$,
$$\sum_{k=0}^{2n}c_{k}x^{k}\le \sum_{k=0}^{\infty}c_{k}x^{k} \le \sum_{k=0}^{2n+1}c_{k}x^{k}$$
To see this, write
$$\sum_{k=0}^{2n}c_{k}x^{k} = 1+\left(|c_{1}|x^{1} - |c_{2}|x^{2}\right) + \ldots + \left(|c_{2n-1}|x^{2n-1}-|c_{2n}|x^{2n}\right) $$ 
which is a sum of positive terms, and hence positive for every $n$. That is, the sequence $S_{n}= \sum_{k=0}^{2n}c_{k}x^{k}$ is increasing and tends to $f(x)$. So $S_{n}\le f(x)$ for all $n$ (and in fact the inequality is strict). Similarly, 
$$T_{n} := \sum_{k=0}^{2n+1}c_{k}x^{k} = 1+|c_{1}|x^{1} - (|c_{2}|x^{2} - |c_{3}|x^{3}) - \ldots - (|c_{2n}|x^{2n}-|c_{2n+1}|x^{2n+1} $$ 
is a decreasing sequence which tends to $f(x)$. So $T_{n}>f(x)$ for all $n$. Your question is the particular case of the lower bound when $n=1$.
