Determine the maximum value of $\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}$ when $x, y, z$ are strictly positive real numbers satisfying $x+y+z=3$.
According to the Cauchy-Schwarz inequality, we have

$$
\sqrt{x} \sqrt{1}+\sqrt{y+1} \sqrt{2}+\sqrt{z+2} \sqrt{3} \leqslant \sqrt{x+y+1+z+2} \sqrt{1+2+3}=6
$$

Equality is achieved when $x=y=z=1$. Therefore, the maximum value is 6.