Proof will be by contradiction, that is, we show that not just one of x, y, z can be negative, nor can just two of them be negative nor all three negative, leaving the only possibility that all three are positive.
If just one of x, y, z is negative then the product xyz would be negative, which is a contradiction.
If all three of x, y, z are negative, then the sum x + y + z would be negative, which again is a contradiction.
That leaves only the possibility that precisely two of the three are negative.
Suppose that, wlog, \(x = -u^{2}, \text{ and }y = -v^{2},\text{ with }z>0.\)
then, since x + y + z > 0, we have \(-u^{2}-v^{2}+z>0, \text{ so }z > u^{2}+v^{2}\dots(1).\)
From the second condition,
\(\displaystyle u^{2}v^{2}-u^{2}z-v^{2}z>0,\text{ so }\:u^{2}v^{2}>z(u^{2}+v^{2})\text{ and }\:\frac{u^{2}v^{2}}{u^{2}+v^{2}}>z\dots(2).\)
From (1) and (2),
\(\displaystyle \frac{u^{2}v^{2}}{u^{2}+v^{2}}>z>u^{2}+v^{2}, \text{ so }\)
\(\displaystyle u^{2}v^{2}>(u^{2}+v^{2})^{2}=u^{4}+v^{4}+2u^{2}v^{2}\)
again a contradiction, forcing the conclusion that all three are positive.
Tiggsy
