Initially the formulas for $A(n),B(n)$, and $C(n)$ are conjectures derived from Table (1.12) near the foot of page 13: by inspection it appears that $A(n)=2^m$, $B(n)=2^m-1-\ell$, and $C(n)=1$, where $n=2^m+\ell$ and $0\le\ell<2^m$. (The table itself was produced by direct calculation using (1.11).)
As the authors point out, this conjecture can be proved by induction, but it’s a bit messy; it really is easier to take a different approach. The key point to bear in mind is that the functions $A(n),B(n)$, and $C(n)$ are completely determined by (1.11), independent of the values of the parameters $\alpha,\beta$, and $\gamma$. There is a whole family $\mathscr{F}$ of functions $f(n)$ defined from them by $$f(n)=A(n)\alpha+B(n)\beta+C(n)\gamma\;,\tag{1}$$ one for each choice of $\langle\alpha,\beta,\gamma\rangle$. We can use any choice of these parameters and associated function $f$ to extract information about the functions $A(n),B(n)$, and $C(n)$.
In this case we start by guessing that the constant function $f(n)=1$ is a member of the family $\mathscr{F}$. In order for that to be the case, there must be parameters $\alpha,\beta$, and $\gamma$ such that
$$\begin{align*}
1&=\alpha\\
1&=2\cdot 1+\beta\\
1&=2\cdot 1+\gamma\;;
\end{align*}$$
this is simply substituting the function $f(n)=1$ into (1.11). These show that if we set $\alpha=1$, $\beta=-1$, and $\gamma=-1$ in $(1)$, we get the function $f(n)=1$. In other words, for all $n$ we have $1=A(n)-B(n)-C(n)$. This is a fact about the functions $A(n),B(n)$, and $C(n)$; we discovered it by looking at a particular function $f(n)$ and discovering that it is the member of $\mathscr{F}$ obtained when we set $\langle\alpha,\beta,\gamma\rangle=\langle 1,-1,-1\rangle$, but it is necessarily true for every choice of parameter values, because the functions $A(n),B(n)$, and $C(n)$ are independent of the parameter values: they are determined strictly by the recurrence (1.11).
Then we guess (or at any rate hope!) that the identity function $f(n)=n$ belongs to $\mathscr{F}$. Substituting that function into (1.11), we see that this would require that
$$\begin{align*}
1&=\alpha\\
2n&=2\cdot n+\beta\\
2n+1&=2\cdot n+\gamma\;,
\end{align*}$$
and setting $\langle\alpha,\beta,\gamma\rangle=\langle 1,0,1\rangle$ clearly makes this true for all $n$. This implies that $n=A(n)+C(n)$ for all $n$.
If we knew $A(n)$, we could now solve for $B(n)$ and $C(n)$. Here it’s easier to start with parameter values that let us focus entirely on $\alpha$ than to guess another member of $\mathscr{F}$. If we set $\alpha=1$ and $\beta=\gamma=0$, (1.11) becomes
$$\begin{align*}
f(1)&=1\\
f(2n)&=2f(n)\\
f(2n+1)&=2f(n)\;.
\end{align*}\tag{2}$$
This function $f(n)$ isn’t a nice polynomial in $n$, but by our choice of parameters we know that it is $f(n)=A(n)$, we already suspect that $A(n)=2^m$, where $2^m\le n<2^{m+1}$, and $(2)$ is simple enough that it seems reasonable to try to prove by induction that $A(n)=f(n)=2^m$.
This is certainly true for $n=1$: in that case $m=0$, and $2^0=1=f(1)$. Suppose that $f(n)=2^m$ for all $n$ such that $2^m\le n<2^{m+1}$. If $2^{m+1}\le k<2^{m+2}$, let $n=\left\lfloor\frac{k}2\right\rfloor$; then $k=2n$ or $k=2n+1$, depending on whether $k$ is even or odd, and $2^m\le n<2^{m-1}$, so by $(2)$ $f(k)=2f(n)=2\cdot2^m=2^{m+1}$. That is, if $f(n)=2^m$ whenever $2^m\le n<2^{m+1}$, then $f(n)=2^{m+1}$ whenever $2^{m+1}\le n<2^{m+2}$, and the desired result follows by induction. If you want an explicit formula for $m$, note that $2^m\le n<2^{m+1}$ iff $m\le\log_2n<m+1$ iff $m=\lfloor\log_2n\rfloor$. Thus, we now know that $A(n)=2^{\lfloor\log_2n\rfloor}$.
Then $C(n)=n-A(n)=n-2^{\lfloor\log_2n\rfloor}$; if $n=2^m+\ell$, where $0\le\ell<2^m$, this is simply $C(n)=\ell$. And $$B(n)=A(n)-C(n)-1=2^{\lfloor\log_2n\rfloor}-(n-2^{\lfloor\log_2n\rfloor})-1=2^m-\ell-1\;.$$
You might wonder what happens if we try to use a function $f(n)$ that isn’t in $\mathscr{F}$ to get information about $A(n),B(n)$, and $C(n)$. The answer is that we won’t be able to find parameters $\alpha,\beta$, and $\gamma$ consistent with $f(n)$. For example, if you try $f(n)=n^2$, you find that (1.11) becomes
$$\begin{align*}
1&=\alpha\\
(2n)^2&=2n^2+\beta\\
(2n+1)^2&=2n^2+\gamma\;,
\end{align*}$$
and this is impossible: there is no constant $\beta$ such that $4n^2=2n^2+\beta$ for every $n\ge 1$. We see immediately that $f(n)=n^2$ simply isn’t in the family $\mathscr{F}$ of functions of the form $f(n)=A(n)\alpha+B(n)\beta+C(n)\gamma$ for functions $A(n),B(n)$, and $C(n)$ satisfying (1.11).
This answer proves the following claims :
Claim 1 : $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\begin{cases}2\lfloor{\sqrt z}\rfloor&\text{if $\{\sqrt z\}\lt\frac 12$ and $\lfloor \sqrt z\rfloor\gt\dfrac{\{\sqrt z\}^2}{1-2\{\sqrt z\}}$}
\\\\2\lfloor{\sqrt z}\rfloor+1&\text{otherwise}
\end{cases}$$
where $\{x\}$ denotes the fractional part of $x$.
Claim 2 : Conjecture C1 is true.
Claim 1 : $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\begin{cases}2\lfloor{\sqrt z}\rfloor&\text{if $\{\sqrt z\}\lt\frac 12$ and $\lfloor \sqrt z\rfloor\gt\dfrac{\{\sqrt z\}^2}{1-2\{\sqrt z\}}$}
\\\\2\lfloor{\sqrt z}\rfloor+1&\text{otherwise}
\end{cases}$$
where $\{x\}$ denotes the fractional part of $x$.
Proof :
Using that $x-1\lt \lfloor x\rfloor \le x$ and AM-GM inequality, we have
$$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{r_m}\right\rfloor+r_m\gt \frac{z}{r_m}+r_m-1\ge 2\sqrt{z}-1\tag1$$
Also, if $\sqrt{z}=n+a$ where $n\in\mathbb Z$ and $0\le a\lt 1$, we have
$$\begin{align}\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1&=\left\lfloor\dfrac{n^2+2an+a^2}{n+1}\right\rfloor+n+1
\\\\&=\left\lfloor n+2a-1+\frac{(1-a)^2}{n+1}\right\rfloor+n+1
\\\\&=2\lfloor\sqrt z\rfloor+\left\lfloor 2a+\frac{(1-a)^2}{n+1}\right\rfloor\end{align}$$
We can say that $2a+\frac{(1-a)^2}{n+1}\lt 2$ always holds since
$$2a+\frac{(1-a)^2}{n+1}\ge 2\implies \frac{(1-a)^2}{n+1}\ge 2(1-a)\implies \frac{1-a}{n+1}\ge 2\implies -a\ge 2n+1$$which is impossible.
Case 1 : If $a\lt \frac 12$ and $n\gt\frac{a^2}{1-2a}$, then we have
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor-1)=2n+2a-1-2n+1=2a\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor-1$$
and
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor$$
and
$$2a+\frac{(1-a)^2}{n+1}\lt 1\iff 2an+2a+1-2a+a^2\lt n+1\iff n\gt\frac{a^2}{1-2a}$$
So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor$$
Case 2 : If $a\lt \frac 12$ and $n\le\frac{a^2}{1-2a}$ ($\iff n\le 2an+a^2$), then we have
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor-1)=2a\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor-1$$
and
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor$$
and
$$2a+\frac{(1-a)^2}{n+1}\ge 1\iff 2an+2a+1-2a+a^2\ge n+1\iff n\le\frac{a^2}{1-2a}$$
For any integer $c$, we have
$$\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+c}\right\rfloor+\lfloor \sqrt z\rfloor+c=\left\lfloor\dfrac{n^2+2an+a^2}{n+c}\right\rfloor+n+c=2n+\left\lfloor 2a+\frac{(a-c)^2}{n+c}\right\rfloor$$
Here, suppose that
$$2a+\frac{(a-c)^2}{n+c}\lt 1$$
Then, we have $$2an+2ac+a^2-2ac+c^2\lt n+c\implies 2an+a^2\lt n+c-c^2$$
$$\implies n\le 2an+a^2\lt n+c-c^2\implies n\lt n+c-c^2\implies c(c-1)\lt 0$$
which contradicts that $c$ is an integer.
So, we see that if $a\lt \frac 12$ and $n\le\frac{a^2}{1-2a}$, then there is no $r$ such that $\lfloor\frac zr\rfloor+r=2\lfloor\sqrt z\rfloor$.
So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor+1$$
Case 3 : If $a\ge \frac 12$, then we have
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor$$
and
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor+1)=2(a-1)\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor+1$$
and
$$2a+\frac{(1-a)^2}{n+1}\ge 1\iff 2an+2a+1-2a+a^2\ge n+1\iff a^2\ge \underbrace{(1-2a)}_{\le 0}n$$which always holds.
So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor+1.\quad\blacksquare$$
Claim 2 : Conjecture C1 is true.
Proof :
We may suppose that $3\le a\le \sqrt z$ from which we have
$$3\le a\le z\implies (3a-z)(a-3)\le 0\implies 3a^2+3z\le az+9a\implies a+\frac za\le \frac z3+3$$
we get
$$a+b=a+\frac za\le \frac z3+3\tag2$$
It follows from Claim 1 that
$$-\min\bigg(\left\lfloor\frac zr\right\rfloor+r\bigg)\le -2\lfloor\sqrt z\rfloor\tag3$$
Finally, from $(2)(3)$, we have
$$(a+b)-\min\bigg(\left\lfloor\frac zr\right\rfloor+r\bigg)\le \frac z3-2\lfloor\sqrt z\rfloor+3.\quad\blacksquare$$
Best Answer
For the k=3 case, the paper is:
http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=0EF1E16177A3F71FB825A92AE8245B5A?doi=10.1.1.46.1946&rep=rep1&type=pdf
general problem:
https://www.jstor.org/stable/43686318?casa_token=flGruEZzecEAAAAA%3A-H4adfhYl4mJ7kTdv61n0VclU2xdXE_OcBOk7qK2ewmko-HbXSdCMfJlZpQVGbNZwglHDQo-iEzoSwfdIPcnfZNmG9FfAyNWaZQsK-BrZoU0t6QwZg&seq=1#metadata_info_tab_contents
Another variant where each person has $l$ lives:
https://link.springer.com/article/10.1007/s00224-011-9343-6