Prove/disprove an inequality $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ Is $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ TRUE in $(0, \frac{π}{2})$ ?
I proceed by taking two functions $f(x) = \sqrt{1 + x}-1 $ and $g(x) =  \frac{x}{2} - \frac{x^2}{8} $. 
Then $f(0) = 0 = g(0),$ while $f(\frac{π}{2}) > g (\frac{π}{2})$. 
Also both functions are strictly increasing in $(0, \frac{π}{2})$. 
I found nothing else. Does it help?
 A: For first, for any $x>0$
$$\sqrt{1+x}<1+\frac{x}{2}\tag{1}$$
is trivial since both terms are positive, but the square of the LHS is $1+x$ while the square of the $RHS$ is $1+x+\frac{x^2}{4}$. So we just need to bound their difference. Notice that:
$$ 1+\frac{x}{2}-\sqrt{1+x}=\frac{\frac{x^2}{4}}{1+\frac{x}{2}+\sqrt{1+x}}\leq\frac{x^2}{8}\tag{2}$$
so the converse inequality holds over $\mathbb{R}^+$:
$$ \sqrt{x+1}\color{red}{>}1+\frac{x}{2}-\frac{x^2}{8}.\tag{3}$$
A: Integration by parts gives
$$ \sqrt{1+x} = 1 +\frac{x}{2} - \frac{x^2}{8} + \int_0^x \frac{(x-t)^2}{2} \frac{3}{8(1+t)^{5/2}} \, dt, $$
and it is easy to see that the integral on the right-hand side is positive for any $x>0$.
A: $$
\begin{align}
\sqrt{1+x}-(1+\frac x2-\frac{x^2}8)
&=\dfrac{1+x-(1+\frac x2-\frac{x^2}8)^2}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{\frac{x^3}{64}(8-x)}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{x^3}{8x+32-\frac{64}{3+\sqrt{1+x}}}
\end{align}
$$
The denominator is monotonically increasing, and at $x=0$, it is $16$. Thus, for $x\ge0$,
$$
\sqrt{1+x}\ge1+\frac x2-\frac{x^2}8
$$
