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Given $f: \mathbb{R} \to \mathbb{R}$ convex, show that: $$ \frac{2}{3}\left(f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right)\right) \leq f\left(\frac{x+y+z}{3}\right) + \frac{f(x) + f(y) + f(z)}{3}.$$

I have tried some ideas, such as transforming it into $$ f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right) - 3f\left(\frac{x+y+z}{3}\right)\\ \leq f(x) + f(y) + f(z) - f\left(\frac{x+y}{2}\right) - f\left(\frac{z+y}{2}\right) - f\left(\frac{x+z}{2}\right) $$ (which graphically seemed intuitive) and using that such an $f$ lies above its tangents, but did not succeed… Ideas are welcome :)

Michael Rozenberg
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Mirabelle
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    To the previous editor of this post: The “functions” tag is used for questions on properties of mappings in a set-theoretical frame, and this question has nothing to do with that. – Ѕᴀᴀᴅ Sep 29 '18 at 11:25
  • Do you mean $\frac{2}{3}\left( f(\frac{x+y}{2}) + f(\frac{z+y}{2}) + f(\frac{x+z}{2}) \right) \le \ldots $ ? Otherwise it is wrong, as $f(x) = 1$ shows. – Martin R Sep 29 '18 at 18:16
  • Yes, i do, sorry ! – Mirabelle Sep 29 '18 at 21:55
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    The inequality in question is called Popoviciu inequality. – szw1710 Sep 29 '18 at 22:37

1 Answers1

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Let $x\geq y\geq z$.

Consider two cases.

  1. $x\geq y\geq\frac{x+y+z}{3}\geq z$.

From here we see that $2y\geq x+z$ and easy to check that $$\left(\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3},z\right)\succ\left(\frac{x+z}{2},\frac{x+z}{2},\frac{y+z}{2},\frac{y+z}{2}\right),$$ which by Karamata gives $$3f\left(\frac{x+y+z}{3}\right)+f(z)\geq2f\left(\frac{x+z}{2}\right)+2f\left(\frac{y+z}{2}\right).$$ Thus, it's enough to prove that $$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right),$$ which is Jensen.

  1. $x\geq \frac{x+y+z}{3}\geq y\geq z$.

This case is similar. Try to kill it by yourself.

Michael Rozenberg
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  • Thanks. I did not know about Karamata's inequality. Seems pretty powerful ! – Mirabelle Sep 29 '18 at 22:00
  • This is strongly connected with so-called majorization and producing Schur-convex sums by convex functions. Not everybody knows it. I think this is rather for specialist in convex functions. – szw1710 Sep 29 '18 at 22:37