Find all triplets $(a, b, c)$ of strictly positive real numbers such that

\[
\left\{\begin{array}{l}
a \sqrt{b}-c=a \\
b \sqrt{c}-a=b \\
c \sqrt{a}-b=c
\end{array}\right.
\]
We notice that $\mathrm{a}=\mathrm{b}=\mathrm{c}=4$ is a solution. We will show that this is in fact the only one.
If two of the three numbers are equal to 4 (say a and b), it is easily verified that the third one is also 4, because $4 \sqrt{4}-\mathrm{c}=4$, so $\mathrm{c}=4$. If one of the three numbers is 4 (say a), then $c \sqrt{a}-b=c$ becomes $2 \mathrm{c}-\mathrm{b}=\mathrm{c}$, so $\mathrm{b}=\mathrm{c}$. Moreover, $\mathrm{a} \sqrt{\mathrm{b}}-\mathrm{b}=4$, which can be rewritten as $(\sqrt{\mathrm{b}}-2)^{2}=0$, so $\mathrm{b}=4$ and $\mathrm{c}=4$. It remains to show that it is impossible for the three numbers to all be different from 4.

If this is the case, then either at least two of the numbers are strictly greater than 4, or at least two are strictly less than 4. We treat these two cases separately.
If at least two of the numbers are $>4$, let's say that c is the smallest of the three numbers. Then $a, b>4$, so $a=a \sqrt{b}-c>2 a-c$, so $c>a>4$, which contradicts the minimality of $c$. Similarly, if at least two of the numbers are $<4$, suppose that $c$ is the largest. Then $a, b<4$, so $a=a \sqrt{b}-c<2 a-c$, so $c<a$, which contradicts the maximality of $c$. The only solution is therefore indeed $(4,4,4)$.