9
$\begingroup$

I refine a famous inequality this is the following :

Let $x,y>0$ then we have : $$x^n+y^n\leq \Big(\frac{x^n+y^n}{x^{n-1}+y^{n-1}} \Big)^n+\Big(\frac{x+y}{2}\Big)^n$$ It's equivalent to : $$x^n+1\leq \Big(\frac{x^n+1}{x^{n-1}+1} \Big)^n+\Big(\frac{x+1}{2}\Big)^n$$ Because it's homogeneous . We can't use AM GM it's too weak so the difficulty is interesting . I try to derivate this but it's a little bit ugly . I have two questions how interpreting this result and how to solve this one variable inequality ?

Thanks a lot

$\endgroup$
  • 2
    $\begingroup$ what are you exactly asking? You want us to prove the inequality you wrote or do you want to find out if it is true? $\endgroup$ – dezdichado Jul 27 at 0:48
  • 1
    $\begingroup$ One of the tags is contest-math. Is this from an ongoing contest? $\endgroup$ – user43208 Aug 2 at 22:51
  • 1
    $\begingroup$ @FatsWallers Your question was taken by Michael Rozenberg to mathoverflow.net/questions/337457 and found quite some interest (and two answers) there. Thus it would be nice if you tell a little more about your question. What is the "famous inequality" you are referring to? Did you conjecture the posted inequality, and if not, where did you find it? $\endgroup$ – Peter Mueller Aug 10 at 16:47
  • 1
    $\begingroup$ Maybe this could explain the motivation: if you look at this question of FatsWallers math.stackexchange.com/questions/3260994/… the right hand side is exactly the inequality asked here for $f(x)=x^{n}$. The author claims "it follows from karamata" - this is not true, for example, $f(x)=x^{3}+x^{2}+100 x$ gives counterexample on $[0,1]$. Nevertheless, apparently by some reasons the inequality still holds for $f(x)=x^{n}, n\geq 1$ and, in a sense, it does refine Karamata's inequality even though vectors do not majorize each other. $\endgroup$ – Paata Ivanishvili Aug 12 at 15:49
  • 1
    $\begingroup$ @PaataIvanishvili Yes, that's very plausible. And right, the answer to the other question is no, even for the simpler example $x^3+ax$ for every positive $a$. $\endgroup$ – Peter Mueller Aug 12 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy