23. Solution: (1) Since $a^2 + b^2 \geq 2ab$, $b^2 + c^2 \geq 2bc$, $c^2 + a^2 \geq 2ac$, and $abc = 1$, we have $a^2 + b^2 + c^2 \geq ab + bc + ca = \frac{ab + bc + ca}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.\\
Therefore $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq a^2 + b^2 + c^2$.\\
(2) Since $a, b, c$ are positive numbers and $abc = 1$, we have\\
$(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 3\sqrt[3]{(a+b)^3(b+c)^3(a+c)^3}$\\
$= 3(a+b)(b+c)(a+c)$\\
$\geq 3 \times (2\sqrt{ab}) \times (2\sqrt{bc}) \times (2\sqrt{ac})$\\
$= 24$.\\
Therefore $(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 24$.