<s>[INST] Triples of positive real numbers $(a,b,c)$ such that $\lfloor a\rfloor bc=3,\; a\lfloor b\rfloor c=4,\;ab\lfloor c\rfloor=5$ Find the all ordered triplets of positive real numbers  $(a,b,c)$ such that: $$\lfloor a\rfloor bc=3,\quad a\lfloor b\rfloor c=4,\quad ab\lfloor c\rfloor=5,$$
where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
 [/INST] Now we have by dividing equations and putting together the ratios we get the ratio $$\frac a{\lfloor a\rfloor}:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$
This shows that $\frac a{\lfloor a\rfloor}\ge\frac53$ since $\frac c{\lfloor c\rfloor}\ge1$. It is checkable that $\frac a{\lfloor a\rfloor}\ge\frac53$ forces $\lfloor a\rfloor=1$, since $a,b,c\ge1$
Similarly, we get $\frac b{\lfloor b\rfloor}\ge\frac54$ This forces $\lfloor b\rfloor\le3$
Now,$$a=\frac a{\lfloor a\rfloor}\ge\frac53$$
we already have $ab\ge\frac{25}{16}$. But $ab\lfloor c\rfloor=5$ so that $\lfloor c\rfloor\le\frac{80}{25}$ whence $\lfloor c\rfloor\le 3$ But if one of $b,c$ is $3$ or more then it forces the other two to be $1$ or less which contradicts either the ratios above or the fact that all of a,b,c are at least 1.
So, $\lfloor b\rfloor=\lfloor c\rfloor\le2$ and $\lfloor a\rfloor=1$
The ratios become
$$a:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$
Case 1: $\lfloor b\rfloor=1$
Then $a:b=4:3$, so $b=\frac{3a}4$. The third equation in data then gives$$\frac{3a^2}4\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=20$. But $a<2$ implies $\lfloor c\rfloor=2$ Now we get $a=\sqrt\frac{10}3$ $b=\frac{\sqrt{30}}4$ Since $\lfloor a\rfloor=1$, we have by the first equation in data that $\frac{\sqrt{30}c}4=3$ That is, $c=\frac25{\sqrt{30}}$
Case 2: $\lfloor b\rfloor=2$
Then $a:b=2:3$, so $b=\frac{3a}2$. The third equation in data then gives$$\frac{3a^2}2\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=10$.
Suppose $\lfloor c\rfloor=1$, we get $3a^2=10$ again and as above using ratios we get the same $a$ and $b$, which contradicts that $\lfloor b\rfloor\ne1$ here! This contradiction gives $\lfloor c\rfloor=2$ and we get $6a^2=10$ and $a=\sqrt\frac53$ But we saw earlier that $a\ge\frac53$. So case 2 is contradictory and we have only 1 solution!
$$(a,b,c)=\left(\sqrt\frac{10}3,\frac{\sqrt{30}}4,\frac25{\sqrt{30}}\right)$$
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