4. Let $x, y, z$ be non-negative numbers such that $x^{2}+y^{2}+z^{2}+x+2 y+3 z=\frac{13}{4}$. Find the minimum value of $x+y+z$.
(1 mark)設 $x 、 y 、 z$ 為非負數, 使得 $x^{2}+y^{2}+z^{2}+x+2 y+3 z=\frac{13}{4}$ 。求 $x+y+z$ 的最小值。
4. $\frac{-3+\sqrt{22}}{2}$
4. Since $x, y, z$ are non-negative we have
$$
\frac{13}{4}=x^{2}+y^{2}+z^{2}+x+2 y+3 z \leq(x+y+z)^{2}+3(x+y+z) .
$$

Solving the quadratic inequality (subject to $x, y, z$ and hence their sum being non-negative) gives $x+y+z \geq \frac{-3+\sqrt{22}}{2}$. Equality is possible when $x=y=0$ and $z=\frac{-3+\sqrt{22}}{2}$. It follows that the minimum value of $x+y+z$ is $\frac{-3+\sqrt{22}}{2}$.

Remark. Intuitively, since the coefficient of $z$ is greater than the coefficients of $x$ and $y$, making $z$ relatively big and $x, y$ relatively small could reduce the value of $x+y+z$. As $x, y, z$ are non-negative it would be natural to explore the case $x=y=0$.