, 2002

Solution:

Note the adjacent figure.
$DP$ and $DQ$ are the parallels through $D$ to $AB$ and $AC$. Since $AD = x$ is the angle bisector of $\alpha$, $AQDP$ is a rhombus, whose side length is denoted by $u$.
From the similarity of triangles $PDC$ and $QBD$ it follows that $\frac{u}{c-u} = \frac{b-u}{u}$, from which one obtains $u = \frac{bc}{b+c}$.
The law of cosines in triangle $AQD$ leads to
$$\begin{aligned} x^{2} = 2u^{2} - 2u^{2} \cos(\pi - \alpha) & = 2u^{2}(1 + \cos \alpha) \\ = 2u^{2}\left(1 + \frac{b^{2} + c^{2} - a^{2}}{2bc}\right) = 2u^{2} & \frac{(b+c)^{2} - a^{2}}{2bc} = \\ & = \frac{bc}{(b+c)^{2}}(a+b+c)(-a+b+c) \end{aligned}$$

With $(b+c)^{2} \geq 4bc$ and $a+b+c = 6$ it follows that $x^{2} \leq 1,5(-a+b+c)$.
Similarly one obtains $y^{2} \leq 1,5(a-b+c)$ and $z^{2} \leq 1,5(a+b-c)$.
From the inequality between the arithmetic and geometric mean of three positive numbers it follows that $\left(\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}}\right)\left(x^{2} + y^{2} + z^{2}\right) \geq 9$ and from this finally
$$\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} \geq \frac{9}{x^{2} + y^{2} + z^{2}} \geq \frac{9}{1,5 \cdot 6} = 1$$
where equality holds only for $x = y = z$.