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I am still trying to find a true statement :-(

Assume that for $x(\alpha,\beta) > 0$, $y(\alpha,\beta) > 0$ and $\alpha > 0$ and and $\beta > 0$ and for all $q>1$ holds $\alpha \leq \frac{\beta(x(\alpha,\beta)^q) + y(\alpha,\beta)^q}{x+y}$. It follows that $\alpha \leq \frac{f(x(f)) +f(y(f))}{x+y}$ holds for all strictly increasing and strictly convex functions f(x) with $f(0) \geq 0$.

Is the above statement true? If yes, can you give a source for it?

My intuition is that $x^q$ with $q→1$ is the flattest possible strictly convex function. Therefore, the statement should be true.

Thank you

Paul
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  • $1\leq (x^q +y^q)/(x+y)$ for all $q > 1$. But $1\leq (f(x) +f(y))/(x+y)$ does not hold for all strictly convex functions, e.g. not for $f(x) = e^{-x}$. – Martin R Sep 14 '19 at 09:00
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    Also this cannot be true because you can scale $f$ by a constant factor. – Martin R Sep 14 '19 at 09:06
  • I really hoped that you are not working today, because I am so emberessed right now by my inability to get my questions right. This time I missed strictly increasing. Is there any option on this page to give someone extra credit for his/her helpfulness? – Paul Sep 14 '19 at 10:41
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    I don't see how the additional condition that $f$ is increasing helps. You can still scale $f$ by a constant factor, so that $\alpha \leq (f(x) +f(y))/(x+y)$ becomes wrong. – Martin R Sep 14 '19 at 10:55
  • Wouldn't this scaling then imply that f(0)<0? – Paul Sep 14 '19 at 14:17
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    No. You can multiply an arbitrary convex function $f$ with a (sufficiently small) positive constant $c$ so that $\tilde f = cf$ does not satisfy $\alpha \leq (\tilde f(x) + \tilde f(y))/(x+y)$. – Martin R Sep 14 '19 at 16:26
  • Absolutely correct. Thank you very much – Paul Sep 15 '19 at 10:52

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