Proof: Let $f(x, y, z)=\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}-\frac{2}{x y z}$, and assume without loss of generality that $x=\max \{x, y, z\} \in\left[\frac{\sqrt{3}}{3}, 1\right]$, then
$$\begin{aligned} & f(x, y, z)-f\left(x, \sqrt{\frac{y^{2}+z^{2}}{2}}, \sqrt{\frac{y^{2}+z^{2}}{2}}\right) \\ = & \frac{\left(x(y+z)^{2}-2 y z\right)(y-z)^{2}}{x y^{2} z^{2}\left(y^{2}+z^{2}\right)} \geqslant 0 . \end{aligned}$$

In fact, $x(y+z)^{2} \geqslant \frac{4 \sqrt{3}}{3} y z$.
Because
$$\begin{aligned} f\left(x, \sqrt{\frac{y^{2}+z^{2}}{2}}, \sqrt{\frac{y^{2}+z^{2}}{2}}\right) & =f\left(x, \sqrt{\frac{1-x^{2}}{2}}, \sqrt{\frac{1-x^{2}}{2}}\right) \\ & =\frac{-3 x+1}{x^{2}(x+1)} \geqslant 9-6 \sqrt{3} \end{aligned}$$

Therefore, equation (43) holds, with equality if and only if $x=y=z=\frac{\sqrt{3}}{3}$.