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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable in $\mathbb{R}$ such that $f'(x)>1$ for each $x>0$.

I want to show that there is $\xi >2$ such that $\frac{f}{x}>\frac{1}{2}$ for each $x>\xi$.

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For that do we have to use the mean value theorem?

But for which interval? I got confused with the $2$ at $\xi>2$.Do we have to consider for each $x>2$ the intervals $[2,x]$ to apply that theorem?

Is yes, then we have that there exists $\xi\in (2,x)$ such that $$\frac{f(x)-f(2)}{x-2}=f'(\xi)>1$$

How do we continue from here?

Or do we consider the intervals $[2,x]$ instead?

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First, prove that $f(x) > x+f(0)$ for $x>0$, using the MVT. From there you easily get to $f(x)> \frac 12x$ for $x>-2f(0)$.

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  • $\begingroup$ Ok! I proved the first part. For the second part do we apply again the MVT? Or do we just use an inequality for $f(0)$ ? $\endgroup$ – Mary Star Sep 6 at 11:14
  • $\begingroup$ Keyword: transitivity of relations $\endgroup$ – amsmath Sep 6 at 11:19
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    $\begingroup$ $$ (1) \ \ \ \ \ \sum_{n[\xi]+1}^{\infty}\frac{1}{f(n)\log^2(n+1)} \\ (2) \ \ \ \ \ \sum_{n=0}^{\infty}\frac{n^2}{(f(x))^n} \ \ \text{ with } x>\xi$$ At (1) I used again the inequality $f(n)>\frac{n}{2} \Rightarrow \frac{2}{n+1}>\frac{1}{f(n)}$ and so $\frac{1}{f(n)\log^2(n+1)}<\frac{1}{(n+1)\log^2(n+1)}$. Using integral criterion we see that the series of the right side converges and so also the left one. $\endgroup$ – Mary Star Sep 6 at 12:55
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    $\begingroup$ At (2) we have for $x>\xi$ that $f(x)>\frac{x}{2}$ and so $$\frac{n^2}{(f(x))^n}<n^2\cdot \left (\frac{2}{x}\right )^n$$ Using the ratio test for the series of $n^2\cdot \left (\frac{2}{x}\right )^n$ using the fact that $\frac{2}{x}<1$ since $x>\xi>2$ we see that the series of the right side converges and so also the left one. Is everythig correct? $\endgroup$ – Mary Star Sep 6 at 12:55
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    $\begingroup$ @MaryStar Yes, this is all correct. Very good! You get an A. ;-) $\endgroup$ – amsmath Sep 6 at 15:33

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