Prove that by the AM-GM inequality,
$$\begin{array}{l} a^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 a\left(\frac{a+b+c+d}{4}\right)^{2} \\ b^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 b\left(\frac{a+b+c+d}{4}\right)^{2} \\ c^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 c\left(\frac{a+b+c+d}{4}\right)^{2} \\ d^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 d\left(\frac{a+b+c+d}{4}\right)^{2} \end{array}$$

Adding and simplifying the above four inequalities, we get
$$a^{3}+b^{3}+c^{3}+d^{3} \geqslant \frac{1}{16}(a+b+c+d)^{3}=\frac{1}{16}$$

Thus,
$$2\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant \frac{1}{8}$$

By the Cauchy-Schwarz inequality,
$$\begin{aligned} a^{2}+b^{2}+c^{2}+d^{2}= & \frac{1}{4}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(1+1+1+1) \geqslant \\ & \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4} \end{aligned}$$

Using the Cauchy-Schwarz inequality again,
$$\begin{aligned} a^{3}+b^{3}+c^{3}+d^{3}= & \left(a^{3}+b^{3}+c^{3}+d^{3}\right)(a+b+c+d) \geqslant \\ & \left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} \end{aligned}$$

Combining (2), we get
$$a^{3}+b^{3}+c^{3}+d^{3} \geqslant \frac{1}{4}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$$

By (1) + (3) $\times 4$, we obtain the desired inequality.