Prove that,
$$\begin{array}{l} \left(a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}\right)^{2} \\ \geqslant\left(a^{\frac{15}{8}}+3 b^{\frac{5}{3}} c^{\frac{5}{8}} d^{\frac{5}{8}}\right)^{2} \\ =a^{\frac{15}{4}}+6 a^{\frac{15}{8}} b^{\frac{5}{8}} c^{\frac{5}{8}} d^{\frac{5}{8}}+9 b^{\frac{5}{4}} c^{\frac{3}{4}} d^{\frac{5}{4}} \\ \geqslant a^{\frac{15}{4}}+15 \sqrt[15]{a^{\frac{15}{4}} b^{15} c^{15} d^{15}} \\ =a^{\frac{15}{4}}+15 a^{\frac{3}{4}} b c d \\ \Rightarrow\left(a^{\frac{15}{8}}+b^{\frac{5}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}\right)^{2} a^{3} \\ \geqslant a^{\frac{27}{4}}+15 a^{\frac{15}{4}} b c d=\left(a^{3}+15 b c d\right) a^{\frac{15}{4}} \\ \Rightarrow \sqrt{\frac{a^{3}}{a^{3}+15 b c d}} \geqslant \frac{a^{\frac{15}{8}}}{a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}} . \end{array}$$

Similarly, $\sqrt{\frac{b^{3}}{b^{3}+15 c d a}} \geqslant \frac{b^{\frac{15}{3}}}{a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}}$,
$$\begin{array}{c} \sqrt{\frac{c^{3}}{c^{3}+15 d a b}} \geqslant \frac{c^{\frac{15}{8}}}{a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}}, \\ \sqrt{\frac{d^{3}}{d^{3}+15 a b c}} \geqslant \frac{d^{\frac{15}{8}}}{a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}} . \end{array}$$

Adding the above four inequalities yields the conclusion.