<s>[INST] 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?
 [/INST] $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{\left(\sqrt{x+1}\right)'''_{x=\xi}x^3}{6}$$ 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!
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