If $x+y+z=xyz$ then is it true that at most one of the numbers can be negative?
My book says this statement is false but I am not able to think of any counter example.
toppr website says this statement is true.
I have found two links on MSE. I am not able to conclude from that either. Here and here.
On
Alternative derivation of counter-example.
Sufficient that two of the three numbers are negative.
Let $x,y,z$ be represented by $1+r,1,1+s,$ respectively.
Then, $r,s$ must be derived to satisfy
$$1 + r + s + rs = 3 + r + s ~: ~r,s < -1 \implies $$
$$rs = 2.$$
This has the easy solution of
$r = s = -\sqrt{2}.$
Then
$$\left(1 - \sqrt{2}\right) \times 1 \times \left(1 - \sqrt{2}\right) = \left(3 - 2\sqrt{2}\right) $$
$$= \left(1 - \sqrt{2}\right) + 1 + \left(1 - \sqrt{2}\right).$$
On
Assume there exists a solution to $x+y+z=xyz$ having one negative number, WLOG let the negative term be $z$, with $x,y$ being positive. Then we know that $xyz$ is a negative number $=-|x||y||z|$
Now multiply each term by $-1$. Now the sum has two negative terms and one positive term, and is equal to the negative of the sum in the first case. Plainly the product is now a positive number $=|x||y||z|$ and the sum is simply the negative of the sum in the first instance, which is $=-(-|x||y||z|)=|x||y||z|$
So a solution with one negative term implies the existence of a solution with two negative terms.
I note by the same argument, if there is a solution with $3$ positive terms. then multiplying by $-1$ affords a solution with $3$ negative terms.
Expanding on Anne Bauval's comment.
$$x+y=z(xy-1)\\\implies z=\frac{x+y}{xy-1}$$
Since, $xy\ne1$,
let $x=-2,y=-1$, thus $z=\frac{-2-1}{2-1}=-3$.
It implies the given claim is incorrect. All the three numbers can be negative simultaneously.
Another example, $x=-\frac12, y=-\frac13$, thus, $z=\frac{-\frac12-\frac13}{\frac16-1}=1$.
It shows exactly two numbers can be negative too.