<s>[INST] 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?
 [/INST] $$
\begin{align}
\sqrt{1+x}-(1+\frac x2-\frac{x^2}8)
&=\dfrac{1+x-(1+\frac x2-\frac{x^2}8)^2}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{\frac{x^3}{64}(8-x)}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{x^3}{8x+32-\frac{64}{3+\sqrt{1+x}}}
\end{align}
$$
The denominator is monotonically increasing, and at $x=0$, it is $16$. Thus, for $x\ge0$,
$$
\sqrt{1+x}\ge1+\frac x2-\frac{x^2}8
$$
</s>