Example 1.36 (Vasile) $a, b, c, x, y, z$ are positive real numbers and satisfy
$$(a+b+c)(x+y+z)=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$
Proof
$$a b c x y z<\frac{1}{36}$$

Proof By the AM - GM inequality, we can obtain
$$\begin{array}{l}
4(a b+b c+c a)(x y+y z+z x)= \\
{\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]\left[(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)\right]=} \\
20-(a+b+c)^{2}\left(x^{2}+y^{2}+z^{2}\right)-\left(a^{2}+b^{2}+c^{2}\right)(x+y+z)^{2} \leqslant \\
20-2 \sqrt{(a+b+c)^{2}\left(a^{2}+b^{2}+c^{2}\right)(x+y+z)^{2}\left(x^{2}+y^{2}+z^{2}\right)}=4
\end{array}$$

Therefore,
$$(a b+b c+c a)(x y+y z+z x) \leqslant 1$$

By the well-known inequality
$$\begin{array}{c}
(a b+b c+c a)^{2} \geqslant 3 a b c(a+b+c) \\
(x y+y z+z x)^{2} \geqslant 3 x y z(x+y+z)
\end{array}$$

Multiplying the two inequalities, we get
$$(a b+b c+c a)^{2}(x y+y z+z x)^{2} \geqslant 9 a b c x y z(a+b+c)(x+y+z)$$

That is, $\square$
$$(a b+b c+c a)^{2}(x y+y z+z x)^{2} \geqslant 36 a b c x y z$$

Thus, from equation (1), we know
$$36 a b c x y z \leqslant 1$$

To achieve equality, we need $a=b=c$ and $x=y=z$, but this contradicts the conditions given in the problem, so equality cannot be achieved. Therefore, the proposition is proved.