Closed form for the limit of the iterated sequence $a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2}$ Is there a general closed form or the integral representation for the limit of the sequence: $$a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}=\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1}=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}$$
in terms of $a_0,b_0,c_0$?
$$L(a_0,b_0,c_0)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n$$
For the most simple case $a_0=b_0$ we have some interesting closed forms in terms of inverse hyperbolic or trigomonetric functions:
$$L(1,1,\sqrt{2})=\frac{1}{\ln(1+\sqrt{2})}$$
$$L(1,1,1/\sqrt{2})=\frac{2 \sqrt{2}}{\pi}$$
$$L(1,1,2)=\frac{\sqrt{3}}{\ln(2+\sqrt{3})}$$
$$L(1,1,1/2)=\frac{3 \sqrt{3}}{2 \pi }$$
$$L(1,1,3)=\frac{\sqrt{2}}{\ln(1+\sqrt{2})}$$
$$L(1,1,1/3)=\frac{2 \sqrt{2}}{3 \arcsin \left( \frac{2 \sqrt{2}}{3} \right)}$$
$$L(1,1,\sin 1)=\frac{2 \cos 1}{\pi -2}$$
$$L(1,1,\sin 1/2)=\frac{2 \cos 1/2}{\pi -1}$$
$$L(1,1,\cosh 1)=\sinh 1$$
$$L(1,1,\cosh 2)=\frac{\sinh 2}{2}$$
These values are obtained by Wolfram Alpha and Inverse Symbolic Calculator and checked with Mathematica.
This result seems familiar to me, but I'm pretty sure I don't know how to obtain a general closed form or even integral representation.

This question is closely related, but the sequence is very different.

Update:
It seems likely that the closed form for the special case $a_0=b_0=1$ is:

$$L(1,1,x)=\frac{\sinh \left(\cosh ^{-1}\left(x\right)\right)}{\cosh ^{-1}\left(x\right)}$$

However, the proof would be nice as well as the more general case.

Thanks to Sangchul Lee I now see that this sequence is exactly Carlson's transformation. Change:
$$a^2=A,\quad b^2=B,\quad c^2=C$$
$$A_{n+1}=\frac{1}{4} (A_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
$$B_{n+1}=\frac{1}{4} (B_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
$$C_{n+1}=\frac{1}{4} (C_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
Accoding to Wikipedia and references wherein, we have:
$$R_F(A_{n+1},B_{n+1},C_{n+1})=R_F(A_n,B_n,C_n)$$
$$R_F(A,B,C)=\frac{1}{2} \int_0^\infty \frac{dt}{\sqrt{(t+A)(t+B)(t+C)}}$$
Which is exactly the 'closed form' I wanted.
 A: Here is an observation: As in your link, if $b_0 = c_0$ then we can prove recursively that $b_n = c_n$ for all $n \geq 0$. Then plugging this to OP's recurrence relation, we find that the sequence $(a_n, b_n)$ satisfies the Schwab-Borchardt recurrence relation
$$ a_{n+1} = \frac{a_n + b_n}{2}, \qquad b_{n+1} = \sqrt{a_{n+1}b_n}. $$
So if we write $(a_0, b_0) = (a, b)$, then the limit is given by the Schwab-Borchardt mean
$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = SB(a_0, b_0)
= \begin{cases}
\dfrac{\sqrt{a^2 - b^2}}{\operatorname{arccosh}(a/b)}, & a > b \\
\dfrac{\sqrt{b^2 - a^2}}{\operatorname{arccos}(a/b)}, & a < b \\
a, & a = b
\end{cases} \tag{*} $$

Anyway, let me write down my failed attempt to mimic Carlson's proof of $\text{(*)}$ so that future me do not repeat this mistake.
Define $I(a,b,c)$ by
$$ I(a,b,c) := \lim_{R\to\infty} \int_{-R}^{R} \frac{dx}{(x+a^2)^{1/3}(x+b^2)^{1/3}(x+c^2)^{1/3}}, $$
where $z^{1/3} = \exp(\frac{1}{3}\log z)$ with the branch cut $-\pi < \arg(z) \leq \pi$. Then we  can check that $I(a, b, c)$ converges. Now using the substitution $x \mapsto 4x + ab + bc + ca$, we find that
$$ I(a, b, c) = I\left( \frac{\sqrt{(a+b)(a+c)}}{2}, \frac{\sqrt{(b+c)(b+a)}}{2}, \frac{\sqrt{(c+a)(c+b)}}{2} \right). $$
This tells us that $I(a_n,b_n,c_n) = \cdots = I(a_0,b_0,c_0)$. If we recall how the AGM is computed from Landen's transformation, this may possibly allow us to analyze $L$ using $I$.
Well, it turns out that the function $I$ has a serious issue, which is that $I$ is identically zero! This can be checked either by using the fact $I(a,a,a) = 0$ or by interpreting $I(a,b,c)$ as a value of the Schwarz–Christoffel mapping.
A: I can show that
the sum of the squares of the terms
decreases at each step.
I think this implies convergence,
but I am not completely sure.
$a_{n+1}
=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ 
b_{n+1}=
\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ 
c_{n+1}
=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}
$
I'm going to play around
and see if anything interesting happens,
with a goal of proving convergence,
maybe.
Multiplying, we get
$a_{n+1}b_{n+1}c_{n+1}
=(a_n+b_n)(a_n+c_n)(b_n+c_n)/8
=(a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2)/8
$
(omitting the "_n")
so that
$\begin{array}\\
\dfrac{a_{n+1}b_{n+1}c_{n+1}}{abc}
&=(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + 2  + \frac{c}{b} + \frac{b}{a} + \frac{c}{a})/8\\
&=(\frac{a}{c} + \frac{c}{a} + \frac{a}{b} + \frac{b}{a}  + \frac{b}{c}+ \frac{c}{b} + 2 )/8\\
&=(\frac{a}{c} -2+ \frac{c}{a} + \frac{a}{b}-2 + \frac{b}{a}  + \frac{b}{c}-2+ \frac{c}{b} + 8 )/8\\
&=(g(\dfrac{a}{c}) + g(\dfrac{a}{b})  + g(\dfrac{b}{c}))/8+1
\qquad\text{where } g(x) = (\sqrt{x}-\dfrac1{\sqrt{x}})^2=x+\dfrac1{x}-2\\
\end{array}
$
Note that 
$g'(x)
=1-\dfrac1{x^2}
$.
Don't know what to do with this,
so I'll try something else.
$\begin{array}\\
a_{n+1}-a
&=\frac{\sqrt{(a+b)(a+c)}}{2}-a\\
&=a(\frac{\sqrt{(1+b/a)(1+c/a)}}{2}-1)\\
\end{array}
$
Again nothing.
Let's try this.
Since
$(a+b+c)^2
\le 3(a^2+b^2+c^2)
$
with equality iff
$a=b=c$,
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
&=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\
&=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\
&=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\
&=(\frac32(a^2+b^2+c^2+2ab+2ac+2bc)-\frac12(a^2+b^2+c^2))/4\\
&=(\frac32(a+b+c)^2-\frac12(a^2+b^2+c^2))/4\\
&\le(\frac32 3(a^2+b^2+c^2)-\frac12(a^2+b^2+c^2))/4\\
&=a^2+b^2+c^2\\
\end{array}
$
Therefore,
if at least two of
$a, b, c$ are distinct,
the sum of their squares
decreases.
This is not a proof of convergence,
but a start.
Let's try a variation on 
this last result.
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
&=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\
&=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\
&=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\
\end{array}
$
so
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
-(a^2+b^2+c^2)
&=(-3a^2-3b^2-3c^2+3ab+3ac+3bc)/4\\
&=-3(a^2+b^2+c^2-ab-ac-bc)/4\\
&=-3(2a^2+2b^2+2c^2-2ab-2ac-2bc)/8\\
&=-3((a-b)^2+(a-c)^2+(b-c)^2)/8\\
\end{array}
$
This shows precisely
how the sum of squares
decreases at each step.
I think that
this implies convergence.
