Đáp án:
$\begin{array}{l}
Theo\,Cô - si:1 + {x^2} \ge 2\sqrt {1.{x^2}} = 2x\\
\Rightarrow \left\{ \begin{array}{l}
1 + {y^2} \ge 2y\\
1 + {z^2} \ge 2z
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{x}{{\sqrt {yz\left( {1 + {x^2}} \right)} }} \le \frac{x}{{\sqrt {yz.2x} }} = \frac{1}{{\sqrt 2 }}.\frac{{\sqrt x }}{{\sqrt {{\rm{y}}z} }}\\
\frac{y}{{\sqrt {xz\left( {1 + {y^2}} \right)} }} \le \frac{1}{{\sqrt 2 }}.\frac{{\sqrt y }}{{\sqrt {xz} }}\\
\frac{z}{{\sqrt {xy\left( {1 + {z^2}} \right)} }} \le \frac{1}{{\sqrt 2 }}.\frac{{\sqrt z }}{{\sqrt {xy} }}
\end{array} \right.\\
\Rightarrow K \le \frac{1}{{\sqrt 2 }}\left( {\frac{{\sqrt x }}{{\sqrt {yz} }} + \frac{{\sqrt y }}{{\sqrt {xz} }} + \frac{{\sqrt z }}{{\sqrt {xy} }}} \right)\\
\Rightarrow K \le \frac{1}{{\sqrt 2 }}.\left( {\frac{{x + y + z}}{{\sqrt {xyz} }}} \right)\\
\Rightarrow K \le \frac{1}{{\sqrt 2 }}\\
\Rightarrow Max\,K = \frac{1}{{\sqrt 2 }} \Leftrightarrow x = y = z = \frac{1}{{\sqrt 3 }}
\end{array}$
