# Does AM-GM Follow From the Convexity of Some Function

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?

• 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 Jan 17, 2021 at 21:00
• @leoli1 That is interesting, thank you. Jan 17, 2021 at 21:15

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 $$f(x)$$.