3. Let $x, y, z$ be positive real numbers such that $x+y+z=3$. Find the maximum value reached by

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

For what values of $x, y, z$ is this maximum achieved?
Solution 1. Consider the vectors $(\sqrt{x}, \sqrt{y+1}, \sqrt{z+2})$ and $(\sqrt{1}, \sqrt{2}, \sqrt{3})$, whose coordinates are all real and positive, whose respective magnitudes are $\sqrt{x+y+z+3}=$ $\sqrt{6}$ and $\sqrt{1+2+3}=\sqrt{6}$, and whose dot product is the expression whose maximum is to be found. By the dot product inequality, the maximum value is equal to the product of the magnitudes of the vectors, which is 6, with equality occurring when both vectors are proportional. In this case, since both vectors have the same magnitude, equality occurs if and only if both vectors are equal, that is,

$$
\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6} \leq 6
$$

with equality if and only if $x=y=z=1$.