The AM-GM for $n = 2$ is $$\frac{x+y}{2} \ge (xy)^{1/2}$$ This is very easy to prove with algebra. However, I am wondering if there is a proof using convexity. That is, can we find a convex function $f:\mathbb{R} \to \mathbb{R}$ so that for some $t = t(x, y)$ between $0$ and $1$ we have $$tf(x) + (1-t)f(y) = \frac {x+y}{2}$$ and $$f(tx + (1-t)y) = (xy)^{1/2}$$

Is there such a proof?

  • 2
    $\begingroup$ Might be not exactly what you are looking for, but the general AM-GM inequality can be proved using the fact that $\log$ is concave, you can find a proof here $\endgroup$
    – leoli1
    Jan 17, 2021 at 21:00
  • $\begingroup$ @leoli1 That is interesting, thank you. $\endgroup$
    – user56202
    Jan 17, 2021 at 21:15

1 Answer 1


No. The first equation implies (by setting $x = y$) that $f(x) = x$ is the identity function, which then forces $t(x,y) = \frac{1}{2}$ for $x \neq y$. Then the second equation cannot be satisfied unless $x = y$.

Even if we only impose the first equation when $x \neq y$, we get the same conclusion. Every convex function $\mathbb{R} \to \mathbb{R}$ is continuous, so $f$ is continuous, so $$f(x) = \lim_{h \to 0} (t f(x) + (1-t) f(x+h)) = \lim_{h \to 0} \frac{x + x + h}{2} = x.$$

To justify the first equality, note that $t f(x) + (1-t) f(x+h)$ lies between $f(x)$ and $f(x+h)$ for all $h$. The intersection of all these intervals $[f(x), f(x+h)]$ (or $[f(x+h),f(x)]$, depending on the value of $h$) is just $\{f(x)\}$ (this is by continuity of $f$). Thus, the limit exists and equals $f(x)$.


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