The means ¯r(a) $\mathcal{\bar{A}}_r(a)$ and (a) $\mathcal{G}(a)$ are of the form

¯ϕ(a)=ϕ−1(∑qϕ(a)) $\mathcal{\bar{A}}_\phi(a) = \phi^{-1}\left(\sum q \phi(a)\right)$,

where ϕ(x) $\phi(x)$ is one of the functions: xr $x^r$ and logx $\log x$ and ϕ−1(x) $\phi^{-1}(x)$ the inverse function. In a nutshell, for ϕ(x)=x,logx $\phi(x) = x, \log x$ and xr $x^r$, ¯ϕ $\mathcal{\bar{A}}_\phi$ reduces to  $\mathcal{A}$,  $\mathcal{G}$ and ¯r $\mathcal{\bar{A}}_r$ respectively.

§ $\S$

In order that

$$\mathcal{\bar{A}}_\psi(a) = \mathcal{\bar{A}}_\chi(a)$$
for all a $a$ and $q$, it is necessay and sufficient that
$$\chi=\alpha\psi+\beta$$ ,
where α $\alpha$ and β $\beta$ are contants and α≠0 $\alpha\neq0$.

Since −ϕ $-\phi$ is a linear function of ϕ $\phi$, and −ϕ $-\phi$ increases if ϕ $\phi$ decreases, we may always suppose, if we please, that the ϕ $\phi$ involved in ¯ϕ(x) $\mathcal{\bar{A}}_\phi(x)$ is an increasing function.

§ $\S$

Suppose that ϕ(x) $\phi(x)$ is continuous in the open interval (0,+∞) $(0,+\infty)$, and that
¯ϕ(ka)=k¯ϕ(a) $\mathcal{\bar{A}}_\phi(ka) = k \mathcal{\bar{A}}_\phi(a)$
for all positive a $a$, $q$, and k $k$. Then $\mathcal{\bar{A}}_\phi(a)$ is ¯r(a) $\mathcal{\bar{A}}_r(a)$.

§ $\S$

log¯rr(a)=rlog¯r(a) $\log \mathcal{\bar{A}}_r^r(a) = r \log \mathcal{\bar{A}}_r(a)$ is a convex function of r $r$.

$\S$

If ϕ(x) $\phi(x)$ is continuous, and there is at least one point of every chord of the curve y=ϕ(x) $y=\phi(x)$, besides the end points of the chord, which lies above or on the curve, then every point of every chord lies above or one the curve, so that ϕ(x) $\phi(x)$ is convex.

§ $\S$

If ψ $\psi$ and χ $\chi$ are continuous and strictly montonic, and χ $\chi$ is increasing, then a necessary and sufficient condition that ¯ψ≤¯χ $\mathcal{\bar{A}}_\psi\le \mathcal{\bar{A}}_\chi$ for all a $a$ and $q$ is that ϕ=χψ−1 $\phi=\chi\psi^{-1}$ should be convex.

§ $\S$

If ϕ(x,y) $\phi(x,y)$ is convex and continuous, then
ϕ(∑qx,∑qy)≤∑qϕ(x,y) $\phi\left(\sum qx,\sum qy\right) \le \sum q \phi(x,y)$

§ $\S$

If

$$\mathcal{A}(ab) \le \mathcal{\bar{A}}_F(a)\mathcal{\bar{A}}_G(b)$$ , in the second the reverse inequality.