59. Find the largest positive real number $a$, such that $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>a$ holds for all positive real numbers $x, y, z$. (1994 Romanian National Training Team Problem)
59. The maximum positive number $a=2$.

On the one hand, let $y=z, x \rightarrow 0^{+}$, then we know $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}} \rightarrow 2$. Therefore, $a \leqslant 2$.
On the other hand, if $a=2$, we will prove that the original inequality holds. Therefore, $a_{\max }=2$.
Without loss of generality, assume $x \leqslant y, z$. To prove $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>2$, it suffices to prove
$$\begin{array}{l}
\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+z^{2}}}+\frac{\sqrt{x^{2}+z^{2}}}{\sqrt{x^{2}+y^{2}}} \Leftrightarrow \\
\frac{1}{\sqrt{y^{2}+z^{2}}}>\frac{x}{\sqrt{x^{2}+z^{2}}\left(y+\sqrt{x^{2}+y^{2}}\right)}+\frac{x}{\sqrt{x^{2}+y^{2}}\left(y+\sqrt{x^{2}+z^{2}}\right)}
\end{array}$$

Since (1) is symmetric with respect to $y, z$, it is sufficient to prove
$$\begin{array}{l}
\frac{1}{2} \cdot \frac{1}{\sqrt{y^{2}+z^{2}}}>\frac{x}{\sqrt{x^{2}+z^{2}}\left(y+\sqrt{x^{2}+y^{2}}\right)} \Leftrightarrow \\
\sqrt{x^{2}+z^{2}}\left(y+\sqrt{x^{2}+y^{2}}\right)>2 x \sqrt{y^{2}+z^{2}} \Leftrightarrow \\
\sqrt{x^{2} y^{2}+y^{2} z^{2}}+\sqrt{x^{4}+x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}}> \\
\sqrt{x^{2} y^{2}+x^{2} z^{2}}+\sqrt{x^{2} y^{2}+x^{2} z^{2}}
\end{array}$$

Since $x \leqslant y, z$, the above inequality clearly holds. Therefore, the original inequality is proved, and the maximum value of $a$ is 2.