$$\begin{array}{l} \sum_{\mathrm{cyc}} \sqrt{2\left(a^{2}+b^{2}\right)}-\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}} \\ =\sum_{\mathrm{cyc}}\left(\sqrt{2\left(a^{2}+b^{2}\right)}-a-b\right)-\left(\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}-2(a+b+c)\right) \\ =\sum_{\mathrm{cyc}} \frac{(a-b)^{2}}{a+b+\sqrt{2\left(a^{2}+b^{2}\right)}} \\ -\frac{\sum_{\mathrm{cyc}}\left(18 a^{3}+27 a^{2} b+27 a^{2} c\right)-8(a+b+c)^{3}}{\left(\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}\right)^{2}+2(a+b+c) \sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}+4(a+b+c)^{2}} \\ =\sum_{\mathrm{cyc}}(a-b)^{2}\left(\frac{1}{a+b+\sqrt{2\left(a^{2}+b^{2}\right)}}\right. \\ \left.-\frac{5 a+5 b+8 c}{\left(\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}\right)^{2}+2(a+b+c) \sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}+4(a+b+c)^{2}}\right) \\ \geqslant \sum_{\mathrm{cyc}}(a-b)^{2}\left(\frac{1}{a+b+1.5(a+b)}\right. \\ \left.-\frac{5 a+5 b+8 c}{\left(\sqrt{3 \sum_{\mathrm{cyc}}(a+b)^{2}}\right)^{2}+4(a+b+c)^{2}+4(a+b+c)^{2}}\right) \\ =\sum_{\mathrm{cyc}}(a-b)^{2}\left(\frac{2}{5(a+b)}-\frac{5 a+5 b+8 c}{\sum_{\mathrm{cyc}}\left(14 a^{2}+22 a b\right)}\right) \\ =\sum_{\mathrm{cyc}} \frac{(a-b)^{2}\left(4 c(7 c+a+b)+3(a-b)^{2}\right)}{10(a+b) \sum_{\mathrm{cyc}}\left(7 a^{2}+11 a b\right)} \geqslant 0, \end{array}$$

Therefore, the original inequality holds.