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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

Remark (@Andreas, 2020-10-25)

  1. This inequality is rather fine-tuned. Consider as a first term on the RHS $$\Big(\frac{x^n+y^n+z\cdot(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+z\cdot(\frac{x+y}{2})^{n-1}}\Big)^n $$ and let $z$ increase from $0$ to $1$. It is easy to see that the increase of $z$ makes the term smaller. Choosing $z=0$ (this question) makes the term "just big enough" for the inequality to be "$\le$". Indeed, for $z=1$, this inverses to "$\ge$", as this post shows. So, fine tuned upper and lower bounds to $x^n + y^n$ are available.

  2. An interesting observation is the following: $$\Big(\frac{x^n+y^n}{x^{n-1}+y^{n-1}} \Big)^n \ge x^n+y^n - \Big(\frac{x+y}{2}\Big)^n \ge \frac{x^n+y^n}{2} .$$ The first inequality is the one under consideration, the second one is an application of Jensen's inequality for two values of the function $x^n$, whereas the inequality between the first and the third expression is a direct application of Slater's inequality* (eq. (2) in this pdf), so we see here a sharpening of Slater's inequality for the function $x^n$.

*Slater ML, A Companion Inequality to Jensen's Inequality. Jour. of Approximation Theory 1981, 32(2):160–166.

Andreas
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    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? – dezdichado Jul 27 '19 at 00:48
  • I think I got a rather cumbersome solution and will type it up sometime tomorrow. – dezdichado Jul 30 '19 at 02:52
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    One of the tags is [tag:contest-math]. Is this from an ongoing contest? – user43208 Aug 02 '19 at 22:51
  • Now there is a solution. – Iosif Pinelis Aug 08 '19 at 14:14
  • @FatsWallers Where you took this problem? – Michael Rozenberg Aug 10 '19 at 13:10
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    @FatsWallers Your question was taken by Michael Rozenberg to https://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? – Peter Mueller Aug 10 '19 at 16:47
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    Maybe this could explain the motivation: if you look at this question of FatsWallers https://math.stackexchange.com/questions/3260994/new-bounds-for-convex-function-of-2-variables 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. – Paata Ivanishvili Aug 12 '19 at 15:49
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    @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 $a$. – Peter Mueller Aug 12 '19 at 22:26

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