$$f(x) = \ln x + \int\limits_0^x {\sqrt {1 + \sin t} \,dt} $$
$$f'(x) = {1 \over x} + \sqrt {1 + \sin x} $$
f' is not differentiable at sin x = $$-$$1
i.e. $$x = 2n\pi - {\pi \over 2},n \in N$$ in the interval (0, $$\infty$$)
$$f''(x) = - {1 \over {{x^2}}} + {{\cos x} \over {2\sqrt {1 + \sin x} }}$$
f'' does not exist for all x $$\in$$ (0, $$\infty$$)
f' exist for x > 0
we have $${1 \over x} + \sqrt {1 + \sin x} < \ln x + \int\limits_0^x {\sqrt {1 + \sin x} dx} $$
because L.H.S. is bounded and R.H.S. is not bounded so $$\exists $$ some $$\alpha$$ beyond which R.H.S. is greater than L.H.S.
i.e. $$|f'(x)| < |f(x)|$$ for all x $$\in$$ ($$\alpha$$, $$\infty$$)
$$|f| + |f'| \le \beta $$ is wrong as f is unbounded while f' is bounded.