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For some convex function $f$, and elements $x_{1},x_{2},x_{3}$ of its domain, show that: $$f(x_{1})+f(x_{2})+f(x_{3})+f\left( \frac{1}{3}(x_{1}+x_{2}+x_{3}\right)\ge \frac{4}{3}\left( f\left(\frac{x_{1}+x_{2}}{2} \right)+f\left(\frac{x_{2}+x_{3}}{2} \right)+f\left(\frac{x_{3}+x_{1}}{2} \right) \right)$$ So far I have that the left hand side is greater than or equal to: $$4f\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)=4f\left(\frac{x_{1}+x_{2}}{6}+\frac{x_{2}+x_{3}}{6}+\frac{x_{1}+x_{3}}{6}\right)$$ by Jensen, which we can say is less than or equal to the right hand side (again, by Jensen). My problem is that this doesn't imply the conclusion above, since $a\ge b\le c$ doesn't tell us anything about the relationship between $a$ and $c$. This makes me think I've made a mistake, which I would be very grateful if someone could spot.

preferred_anon
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2 Answers2

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The following function $f \colon \mathbb R^2 \to \mathbb R$ is a counterexample.

Consider the equilateral triangle $T$ given by the points $(\cos\varphi_i,\sin\varphi_i)$ for $\varphi_1=0$, $\varphi_2=\frac{2\pi}3$, $\varphi_3=\frac{4\pi}3$. (To be more precise, we define the set $T$ only as the boundary of this triangle, and not the interior.)

We put $f(a,b)=\alpha$ for the points belonging to $\alpha T$. (The triangle, which is $\alpha$-times larger.)

This function is convex. (It is the Minkowski functional of the triangle.)

But if we take $x_i=(\cos\varphi_i,\sin\varphi_i)$ for $i=1,2,3$ we have $f(x_i)=f\left(\frac{x_i+x_j}2\right)=1$ and $f\left(\frac{x_1+x_2+x_3}3\right)=0$. So the above inequality now becomes $3\ge \frac43\cdot 3=4$, which is false.

The case of functions $f \colon \mathbb R\to\mathbb R$ (which probably was the original intention of this question) is solved as Problem 431 in the book Putnam and Beyond by Razvan Gelca, Titu Andreescu p.526.

Martin Sleziak
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  • I did not expect that; no wonder I couldn't prove it! – preferred_anon Sep 29 '12 at 16:36
  • To be honest, I have some trouble finding a counterexample for function $\mathbb R\to\mathbb R$. (I am not even sure there is one.) But even if the inequality is true for real function, at least we see from this example, that some property which is no longer true in $\mathbb R^2$ must be used. – Martin Sleziak Sep 29 '12 at 16:39
  • Perhaps the domain should have been $\mathbb{R}$? Regardless, the question should have been straightforward, and I think it's very odd that I'm stuck. I still think it's more likely that I've made a mistake than that the inequality is false. – preferred_anon Sep 29 '12 at 16:41
  • @DanielLittlewood I found a text which contains this problem. (See the links I've added to my post.) Indeed, the problem is formulated there for functions defined on the real line. – Martin Sleziak Sep 29 '12 at 20:15
  • Thank you! It was attributed to Andreescu in my textbook but I couldn't find the solution - it was not so specific there. I can safely say that I would never have got that far. – preferred_anon Sep 29 '12 at 20:19
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Use these:

$f\left(\displaystyle\frac{x_i+x_j}2\right) \le \displaystyle\frac{f(x_i)+f(x_j)}{2}$, and

$f\left(\displaystyle\frac{y_1+y_2+y_3}3\right) \le \displaystyle\frac{f(y_1)+f(y_2)+f(y_3)}{3}$.

This applied to $y_1=\displaystyle\frac{x_2+x_3}2$,.. we also have

$f\left(\displaystyle\frac{x_1+x_2+x_3}3\right) \le \displaystyle\frac{f\left(\frac{x_1+x_2}2\right) + f\left(\frac{x_2+x_3}2\right) + f\left(\frac{x_3+x_1}2\right)}{3} \le\displaystyle\frac{f(x_1)+f(x_2)+f(x_3)}{3}$

Berci
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