If $x, y, z$ aren't sides of a triangle then result is obvious. So we can let $x=a+b, y=b+c, z=c+a$ for positive reals $a, b, c$ and it remains to prove that one of $2b(a+b), 2c(b+c), 2a(c+a)$ is less than or equal to $1$ given $a+b+c=1.5$. But from AM-GM we have \[2b(a+b)\cdot 2c(b+c)\cdot 2a(c+a) = 8abc(a+b)(b+c)(c+a) \leq\frac{8^2}{27^2}\cdot (a+b+c)^6=1\]so $\min{(2b(a+b), 2c(b+c), 2a(c+a))}\leq 1$.

If we can show \[x(x-y-z)y(y+z-x)z(z+x-y)\leq\left(\frac{x+y+z}3\right)^6\]then we're done. It is equivalent to show : \[7\sum_\text{cyc}x^3-6\sum_\text{sym}x^2y+15xyz\geq0\]But by arrangement ineq : $2\sum_\text{cyc}x^3\geq\sum_\text{sym}x^2y$ So it suffices to prove \[5\left(\sum_\text{cyc}x^3-\sum_\text{sym}x^2y+3xyz\right)\geq0\]Which is a Schur's inequality. So this problem can be generalized to : One of $x(x+y-z),y(y+z-x),z(z+x-y)$ should be less than or equal $\left(\frac{x+y+z}3\right)^2$

Let's assume to the contrary: $x (x+y-z) = x (x+y+z - 2z) = x(3-2z)>1$ then $3-2z >\frac{1}{x}$ $y(y+z-x) = y(3-2x)>1$ then $3-2x >\frac{1}{y}$ $z(z+x-y) = z(3-2y)>1$ then $3-2y >\frac{1}{z}$ If we sum all $9 -2(x+y+z) = 3 > \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ Then by $AM-HM$: $\frac{x+y+z}{3} = 1 \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$ Then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 3$ CONTRADICTION. Then at least one of them is less or equal to $1$.

Let $A=x(x+y-z), B=y(y+z-x), C=z(z+x-y)$. Assume that $A, B, C>1$. We have $x+y-z, y+z-x, z+x-y>0$. By AM-GM: $ABC=xyz(x+y-z)(y+z-x)(z+x-y)\leq(\frac{x+y+z}{3})^3 (\frac{x+y-z+y+z-x+z+x-y}{3})^3=1$ Therefore $A, B, C$ can't all be greater than 1 or at least one number is less or equal 1.