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?

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 $x>0$.
