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Prove by induction that: $$(y-x)x^n \leq \frac{y^{n+1}-x^{n+1}}{n+1} \leq (y-x)y^n\ . $$


As a hint, the professor told us to use the following expression that we had previously proven: $$\sum_{i=0}^n{x^i}= \frac{1-x^{n+1}}{1-x} $$

I already tried several things, but I can’t manage to get to the solution.

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closed as off-topic by Saad, John Omielan, user21820, RRL, Alexander Gruber Mar 24 at 2:54

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    $\begingroup$ I suppose you assume that $y>x$? $\endgroup$ – Yanko Mar 22 at 13:33
  • $\begingroup$ It definitely holds for $n=0$. $\endgroup$ – Wuestenfux Mar 22 at 13:34
  • $\begingroup$ Yes sorry. It’s for 0<x<y $\endgroup$ – Facu50196 Mar 22 at 13:34
  • $\begingroup$ And what exactly have you tried? $\endgroup$ – Saad Mar 22 at 13:35
  • $\begingroup$ Were you really told to use induction and to use the result you had previously proved (presumably by using induction)? I can see no need to use both. You could use induction, but this would probably duplicate the proof of the previous result. You can use the previous result directly (i.e. without using induction again) by observing that if $x < 1$, then the left hand side is less than $n+1$ and greater than $(n+1)x^n$. $\endgroup$ – Calum Gilhooley Mar 23 at 15:37
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I'm not quit sure how to use the hint given by the professor. But here's one way to do this:

Recall that $$(y^{n+1}-x^{n+1}) = (y-x)(y^n+y^{n-1}x+....+yx^{n-1}+x^{n})$$

Now assuming that $x<y$ you can replace all the $y$'s with $x$'s in the second multiple1 and get that $$y^{n+1}-x^{n+1} \geq (y-x) (n+1)x^n$$ Divide by $n+1$ and you get the first inequality. For the other one replace the $x$'s with $y$'s.


1.Edit It's possible that you need to prove this part by induction. That is, prove by induction that $$(n+1)x^n \leq y^n+y^{n-1}x+...+yx^{n-1}+x^n$$

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  • $\begingroup$ Thanks!! I’ll try that $\endgroup$ – Facu50196 Mar 22 at 13:58

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