division with the same probability. That is, the following continuations until the next division are equally probable:

-   $a$, $b$, $(a+b)/x$.

-   $a$, $b$, $(a+b)/x$, $(a+(x+1)b)/x$, $(2a+(x+2)b)/x$.

-   $a$, $b$, $(a+b)/x$, $(a+(x+1)b)/x$, $(2a+(x+2)b)/x$, $(3a+(2x+3)b)/x$.

If $b>a$, then the maximum of the last two terms is: $b$, $(2a+(x+2)b)/x$, and $(3a+(2x+3)b)/x$ correspondingly. Adjusting for the fact that $a > b\frac{1}{1+x}$ (see Lemma \[thm:ratio\]), the maximum of the last two terms is at least $b$, $b((x+2)+2/(x+1))/x$, and $b((2x+3)+3/(x+1))/x$ correspondingly. We want to estimate the ratio of the maximum of the last two terms to $\max\{a,b\}$. Counting probabilities we get the following lower bound for the ratio $$1^{1/3}(((x+2)+2/(x+1))/x)^{1/3}(((2x+3)+3/(x+1))/x)^{1/3}= 1.51023.$$

If $b \leq a$, then the maximum of the last two terms is at least $(a+b)/x$, $(2a+(x+2)b)/x$, or $(3a+(2x+3)b)/x$, respectively. Using $b > a\frac{2+x}{(1+x)x}$ from Lemma \[thm:ratio\], we get that the maximum of the last two terms is at least $a\frac{2+2x+x^2}{(1+x)x^2}$, $a\frac{2x^2+6x+4}{(1+x)x^2}$, or $a\frac{5x^2+10x+6}{(1+x)x^2}$, respectively. Counting probabilities, we get that the maximum is multiplied by at least $$\frac{2+2x+x^2}{(1+x)x^2}^{1/3}\frac{3x^2+6x+4}{(1+x)x^2}^{1/3}\frac{5x^2+10x+6}{(1+x)x^2}^{1/3}=0.453822.$$

Notice that if we have 3 or 4 terms until the next division, then the last two terms before the division are in the increasing order. That means the case when $b \leq a$ is at least 