Suppose $f$ is a continuous function on $[0, \infty )$, differentials on $(0, \infty)$, such that $f(0)=1$ and $f'(x)> \frac{1}{2\surd (x+1)} \forall x>0$. Show that $f(x)> \surd (x+1)$.
Rolle's theorem and the mean value theorem are my only leads .
Can an expert provide Hints, tips, suggestions, to a solution?
On
To show the statement in your question is true, first prove that if $f(x)$, $g(x)$ are integrable functions on an interval $[a,b]$, with $f(x) < g(x)$, then $\int_a^b f(x) \, dx < \int_a^b g(x) \, dx$. I would do this using Riemann sums.
To apply this to your particular case, take $f'(t)$ and $\frac{1}{2 \sqrt{t+1}}$ to be your functions which you know are integrable (why?). Then let $[0,x]$ be your interval.
Let $g(x) = f(x) - \sqrt{x + 1}$. Then you have $g(0) = 0$ and $g'(x) > 0$ for all $x > 0$, so...