I claim the answer is $4$. Notice that, if we take $b=c=3/2$, and $a=2+4\sqrt{2}/3$, we achieve $4$. Now, suppose without loss of generality $a\geqslant b\geqslant c$. Now, if $c\geqslant 2$, then for $a=b=c=2$, we have all of the given numbers equal to one, and for $a>2$, we have $(\sqrt{ab}-1)(\sqrt{ac}-1)(\sqrt{bc}-1)>1$, a contradiction. Thus, $c<2$. We now claim to achieve at least four, $c>1$ must hold. Indeed, suppose $c\leqslant 1$. Then, $c-b/a$ and $c-a/b$ are both less than one. Moreover, $b-a/c\leqslant b-a<0$, and thus at most three is achievable. Hence, we restrict ourselves to $1<c<2$. Now, $c-a/b<1$, and thus at most five is achievable. Suppose $5$ is achievable. In particular, we have $c-b/a>1$, $b-a/c>1$, and $a-b/c>1$. Consequently, $bc>a+c$ and $ac>a+b$ are obtained. Now, note that $b>2$ as otherwise $b-a/c>1$ would yield a contradiction. Hence, $a>b>2>c$ is obtained. Now, notice $c-b/a>1$ yields $ac>a+b$. Moreover, $b-a/c>1$ yields $bc>a+c$. Hence, $ac>a+b>2\sqrt{ab}$ and $bc>a+c>2\sqrt{ac}$, giving $abc^2>4a\sqrt{bc}$, that is, $bc^3>16$. Since $c^2<4$, we then get $bc>4$. Finally, $ab>ac>bc>4$ holds. But this yields a contradiction with $(\sqrt{ab}-1)(\sqrt{bc}-1)(\sqrt{ca}-1)=1$.

WLOG suppose $a\geq b\geq c$. Then we have $c\leq2$, because if $c>2$, then $a,b,c>2$. But this gives us a direct contradiction, since $$1=(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)>(\sqrt{4}-1)^3=1$$So we cannot have $c-\frac {a}{b}>1, c-\frac {b}{a}>1$ same time. ($c-\frac {a}{b}>1, c-\frac {b}{a}>1\Rightarrow 2c>2+\frac{a}{b}+\frac{b}{a}\geq4.$) Now one out of the way, we will finally prove that $b-\frac{a}{c}>1$ is not possible too. Suppose on the contrary and let $b-\frac{a}{c}>1$. Then $a\geq b>2$, so we have $$1=(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca} -1)>1.(\sqrt {bc}-1)(\sqrt {ca} -1)\Rightarrow (\sqrt {bc}-1)(\sqrt {ca} -1)<1$$$$c\sqrt{ab}+\sqrt{c}(-\sqrt{a}-\sqrt{b})+1<1\Rightarrow \sqrt{c}<\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}$$ So we have $1<b-\frac{a}{c}<b-\frac{a}{\frac{(\sqrt{a}+\sqrt{b})^2}{ab}}=b-\frac{a^2b}{(\sqrt{a}+\sqrt{b})^2}$. $$\frac{a^2b}{(\sqrt{a}+\sqrt{b})^2}<b-1\Rightarrow a^2<\frac{b-1}{b}(\sqrt{a}+\sqrt{b})^2=(1-\frac{1}{b})((\sqrt{a}+\sqrt{b})^2)\leq (1-\frac{1}{a})(2\sqrt{a})^2=(1-\frac{1}{a})4a=4a-4\Rightarrow$$$$a^2-4a+4<0\rightarrow (a-2)^2<0~~\text{Contradiction.}$$So we can have at most $4$ of them greater than $1$. And the triplet $(\frac{36}{5},\frac{5}{4},\frac{5}{4})$ satisfies the $4$ mark so we are done.

It is easy to see that $(\sqrt{ab}-1)^2 \geq (a-1)(b-1)$ from $A.M-G.M$ Then, we get $(a-1)^2(b-1)^2(c-1)^2 \leqslant 1$ so $(a-1)(b-1)(c-1) \leqslant 1$ Let's look at $a-\frac {b}{c}, b-\frac {c}{a}, c-\frac {a}{b}$ If all three of these numbers are $>1$, then $a-1 > \frac{b}{c}$, $ b-1 > \frac{c}{a}$, $ c-1 > \frac {a}{b}$ and by multiplying these three inequalities we get $(a-1)(b-1)(c-1)>1$, a contradiction. so among these three numbers, at most 2 of them are $>1$. Doing the same thing to the other 3 numbers we get the same result. So at most 4 of the numbers can be $>1$. If we take $b=c=3/2$, and $a=2+4\sqrt{2}/3$, we get the desired result.

CinarArslan wrote: $a, b, c$ are positive real numbers such that $$(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)=1$$At most, how many of the numbers: $$a-\frac {b}{c}, a-\frac {c}{b}, b-\frac {a}{c}, b-\frac {c}{a}, c-\frac {a}{b}, c-\frac {b}{a}$$can be bigger than $1$? W.L.O.G $$a \geq b \geq c \implies ab \geq ac \geq bc$$if all are greater than $1$. $$c - \frac{a}{b} > 1 \implies c>2 \implies a,b,c > 2 \implies ab,ac,bc > 4 $$Contradiction. if $5$ is greater than $1$.in this case only $c - \frac{a}{b} \leq 1$ according to the above case. conditionally $\sqrt {ab}-1 \geq 1 \implies ab \geq 4$ $$c - \frac{b}{a} >1 \implies ac >a+b \geq 2\sqrt {ab} \geq 4$$$$b - \frac{a}{c} >1 \implies bc > a+c \geq 2\sqrt {ac} \geq 4$$then $ab,ac,bc > 4 \implies$ contradiction. Now let’s give 4 examples. $(\frac{36}{5},\frac{5}{4},\frac{5}{4})$