Finding a constant such that this function is positive Given $0<r<1$, is it possible to find $c>0$ such that the function $\left( \frac{x}{2}-1  \right)^{1+2r}- \frac{1}{c}x^{1+2r}+c$ is positive for $x>2$?
I think that this is true because I have maken some experiments for certain values of $r$, but I would like to find a rigurous proof.
 A: The question is equivalent to asking: given $1<s<3$ is there a $c>0$ such that:
$$f(y) := y^s - \frac{2^s}{c}(y+1)^s + c > 0 $$
for all $y>0$ (I just changed the notation slightly).
Let's take $c$ big enough so that $f(1)>0$ and $f'(y)>0$ for all $y\geq1$ (check that we can do that). Thus $f(y)>0$ for all $y\geq1$.
We are now looking for the extremum (minimum) of $f(y)$ on the interval $[0,1]$. It will be $0,1$ or the zero of the derivative of $f$. If $f$ is strictly positive at all of these three points, we get the thesis.
For $c$ big enough $f(0) > 0$ and we already have $f(1) > 0$.
Now find $y_0$ such that $f'(y_0) = 0$. This point $y_0$ is a function of $c$, let's call it $g(c)$.
Now notice that $f(y_0)>0$ for $c$ big enough. That's because $f(g(c))>0$ for $c$ big enough.
Thus we have proven that for $c$ big enough $f(y)>0$ for all $y\geq0$.
Note that we don't need the assumption $s<3$.
A: $\newcommand{\Function}[0]{\displaystyle{\left(\dfrac{x}{2} - 1\right)^{1 + 2r} - \dfrac{1}{c}x^{1+2r} + c}}$Given you're looking to show that $\displaystyle{\exists c > 0}$ such that $\Function > 0$, why not employ limits to exhibit that a large enough $c$, will always result in $\Function > 0$;
$$\lim\limits_{c \to \infty}\Function \to 0 + +\infty = +\infty$$
Moreover, to strengthen the above claim (limit), one can even prove $\left(\dfrac{x}{2} - 1\right)^{1 + 2r} > 0$ and $x^{1+2r} >0, \space \forall x > 2$ and hence our limit w.r.t to $c$ is not affected by $x$ or $r$. Overall, it comes to showing that $\left(\dfrac{x}{2} - 1\right)^{1 + 2r} + c >> \dfrac{1}{c}x^{1+2r}$ as $c \to \infty$
