We know that the arithmetic-geometric mean $AGM(a,b)$ of $a$ and $b$ defined as

$$2a_1=a+b$$ $$b^2_1=ab$$

$$2a_n=a_{n-1}+b_{n-1}$$ $$b^2_n=a_{n-1}b_{n-1}$$

$AGM(a,b)=\lim\limits_{n\to \infty} a_{n}=\lim\limits_{n\to \infty} b_{n}$

$$AGM(a,b)=\frac{\pi}{4}\frac{a+b}{K(\frac{a-b}{a+b})}$$

where $K(m)$ is the complete elliptic integral of the first kind:

$$\int _0^{\frac{\pi}{2}} \frac{1}{\sqrt{a^2 \cos^2(\theta) + b^2 \sin^2(\theta)}} \, d\theta = \int _0^{\frac{\pi}{2}}\frac{1}{\operatorname{AGM}(a,b)} \, d\theta = \frac{\pi}{2 \,\operatorname{AGM}(a,b)}$$

I thought If we use three terms as shown below how it can be named?

$$3a_1=a+b+c$$ $$3b^2_1=ab+ac+bc$$ $$c^3_1=abc$$

$$3a_n=a_{n-1}+b_{n-1}+c_{n-1}$$ $$3b^2_n=a_{n-1}b_{n-1}+a_{n-1}c_{n-1}+b_{n-1}c_{n-1}$$ $$c^3_n=a_{n-1}b_{n-1}c_{n-1}$$

Do they have the limit that all is equal and if yes how to find it?

$x=F(a,b,c)=\lim\limits_{n\to \infty} a_{n}=\lim\limits_{n\to \infty} b_{n}=\lim\limits_{n\to \infty} c_{n} ?$

Does anybody know how to express such this kind of triple mean with known functions ? Can we express $x$ as elliptic integrals?

Thanks for answers

EDIT:

I tested for $a=1$ ,$b=2$,$c=3$

After just 2 iterations , I got $a_2,b_2,c_2 \approx 1.92$

Seems that the results quickly go to a limit .