14. Find the smallest number $k$ such that for all real numbers $x, y$ and $z$
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2} \leq k\left(x^{4}+y^{4}+z^{4}\right) \text {. }
$$
Answer: $k=3$
Solution: Note that
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2}=x^{4}+y^{4}+z^{4}+2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2} .
$$

Using the AM-GM Inequality, we find that
$$
2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2} \leq 2\left[\left(x^{4}+y^{4}\right) / 2\right]+2\left[\left(x^{4}+z^{4}\right) / 2\right]+2\left[\left(y^{4}+z^{4}\right) / 2\right] \text {, }
$$
which when combined with the previous equation results to
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2} \leq 3\left(x^{4}+y^{4}+z^{4}\right) \text {. }
$$

Therefore $k=3$.