Zero uh? No

Clearly $P=0$ when $x=y$ . Therefore $P_{\text{max}}\ge 0$ .

$P$ is cyclic. Wlog $x=\max\{x,y,z\}$ and let $z\neq x\neq y\neq z$ . Two cases:

• $x>y>z$ . Then $P<0$ , so maximum can't be reached.

• $x>z>y$ . Let $(x,y,z)=(y+m+n, y, y+m)$ for some $m,n>0$ .

Then $P=mn(m+n)$ and it's given in the problem statement that $x+y+z=3$ , so $3y+2m+n=3$ .

$P$ is independent of $y$ , so to maximize $P$ let $y=0$ .

Then $P=m(3-2m)(3-m)=2m^3-9m^2+9m$ , which is a polynomial. Maximum value (if it exists) must occur when $P'=0$ . (We have $m>0$ . No endpoints, so $P'$ must equal $0$ . E.g., $Q(x)=x$ is also a polynomial and the maximum value at $x\in[2;5]$ is when $x=5$ , which is an endpoint, and $Q'=1\neq 0$ )

$P'=6m^2-18m+9=0\iff m=\frac{3\pm\sqrt{3}}{2}$ .

If $m=\frac{3+\sqrt{3}}{2}$ , then $P<0$ , so maximum can't be reached.

If $m=\frac{3-\sqrt{3}}{2}$ , then $P=\frac{3\sqrt{3}}{2}\approx\boxed{2.598}$ , which is indeed obtainable when $(x,y,z)=\left(\frac{3-\sqrt{3}}{2}+\sqrt{3},0,\frac{3-\sqrt{3}}{2}\right)$ .