<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] 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 $\mathbb{R}^+$:
$$ \sqrt{x+1}\color{red}{>}1+\frac{x}{2}-\frac{x^2}{8}.\tag{3}$$
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