65. Proof: Let $\sqrt{\frac{v w}{u}}=x, \sqrt{\frac{u w}{v}}=y, \sqrt{\frac{u v}{w}}=z$, then the original proposition is equivalent to: $x, y, z \in \mathbf{R}^{+}$, and $y z+z x+x y+x y z=4$, then
$$x+y+z \geqslant y z+z x+x y$$

The proof of equation (1) can be found in Exercise 64 (1) of this chapter.
Attachment: The solution provided by the National Training Team for this problem (provided by the problem group).
Proof: Let $u=x y, v=y z, w=x z, x, y, z$ be positive real numbers, then the condition becomes
$$x y z+x y+y z+z x=4$$

The conclusion becomes
$$x+y+z \geqslant x y+y z+z x$$

Since $x, y, z$ must have two numbers on the same side of 1, without loss of generality, assume $y, z$ are on the same side of 1, then
$$(y-1)(z-1) \geqslant 0$$

Thus,
$$x(y z+y+z)=4-y z>0$$

From this, we get
$$x=\frac{4-y z}{y z+y+z} \leqslant \frac{4-y z}{y z+2 \sqrt{y z}}=\frac{(2+\sqrt{y z})(2-\sqrt{y z})}{\sqrt{y z}(\sqrt{y z}+2)}=\frac{2-\sqrt{y z}}{\sqrt{y z}}$$

Therefore,
$$(x+1) \sqrt{y z} \leqslant 2$$

Notice that we need to prove
$$\begin{array}{l} \text { Equation (※) } \Leftrightarrow x-x z-x y+x y z \geqslant y z-y-z+x y z \Leftrightarrow \\ x(1-y)(1-z) \geqslant y z(x+1)-y-z \end{array}$$

Next, we prove Equation (※※). By Equation (1), the left side $\geqslant 0$; by Equation (2),
the right side $\leqslant \sqrt{y z} \cdot 2-y-z=-(\sqrt{y}-\sqrt{z})^{2} \leqslant 0$ hence Equation (※※) holds.