4. Rewrite $a, b, c, d$ as $a^{4}, b^{4}, c^{4}, d^{4}$ respectively, then the original expression becomes
$$\begin{array}{c} \sum_{\text {cyc }} \frac{1}{2 a^{8}+a^{5} b c d+a^{2} b^{2} c^{2} d^{2}} \geqslant \frac{1}{a^{2} b^{2} c^{2} d^{2}} . \\ \text { Let } b=a+x, c=a+x+y, d=a+x+y+z \text {, substitute into } \\ \sum_{\text {cyc }}\left(2 b^{8}+b^{5} c d a+a^{2} b^{2} c^{2} d^{2}\right)\left(2 c^{8}+c^{5} d a b+a^{2} b^{2} c^{2} d^{2}\right)\left(2 d^{8}+d^{5} a b c\right. \\ \left.+a^{2} b^{2} c^{2} d^{2}\right) a^{2} b^{2} c^{2} d^{2}-\prod_{\text {cyc }}\left(2 a^{8}+a^{5} b c d+a^{2} b^{2} c^{2} d^{2}\right), \end{array}$$

Prove that this expression $\geqslant 0$.