Mean value thorem - Showing $\sqrt{1+x} < 1+\frac{x}{2}$ for $x>0$ So as the title states I have to show $\sqrt{1+x} < 1+\frac{x}{2}$ for $x>0$.
This is a example from the book which has to be explained for me, i'm having a hardtime understanding the proof. I do however understand the concept of MVT.
So the rest of the solution looks like the following:

If $x>0$, apply the Mean-Value Theorem to $f(x)= \sqrt{1+x}$ on the interval $[0,x]$. There exist $c\in [0,x]$ such that
  $$\frac{\sqrt{1+x}-1}{x}=\frac{f(x)-f(0)}{x-0}=f'(c)=\frac{1}{2\sqrt{1+c}}<\frac{1}{2}
$$
  The last inequality hold because $c>0$. Mulitiplying by the positive number $x$ and transposing the $-1$ gives $\sqrt{1+x} <1+\frac{x}{2}$,

So I am not sure why he(the author) choose $\frac{1}{2}$, it seems a little arbitrary to me. My guess is that you're allowed to pick a number for the derivative of $c$ which suits the cause/solution best, as long as it's $0<c<x$. I'm not sure though.
All help would be greatly appriciated!

Just square both sides.
$(1+x/2)^2
= 1+x+x^2/4
\gt 1+x$
unless $x = 0$.
Also works for
$(1+x/n)^n
\gt 1+x$
so
$1+x/n
\gt \sqrt[n]{1+x}$
(in this case, use
Bernoulli's inequality).
