Prove that because $a^{6}+b^{6} \geqslant a^{4} b^{2}+a^{2} b^{4}$, we have $2\left(a^{6}+b^{6}\right) \geqslant\left(a^{4}+b^{4}\right)\left(a^{2}+b^{2}\right)$, so
$$\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{4}+b^{4}}{2}\right) \leqslant \frac{a^{6}+b^{6}}{2}$$

For any $a, b$, we have $a^{4}+b^{4} \geqslant a^{3} b+a b^{3}$, so
$$2\left(a^{4}+b^{4}\right) \geqslant\left(a^{3}+b^{3}\right)(a+b)$$

So
$$\left(\frac{a+b}{2}\right)\left(\frac{a^{3}+b^{3}}{2}\right) \leqslant \frac{a^{6}+b^{6}}{2}$$

From (1) and (2), we get
$$\left(\frac{a+b}{2}\right)\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{3}+b^{3}}{2}\right) \leqslant \frac{a^{6}+b^{6}}{2}$$