5. For all positive real numbers $a, b, c, d$,
$$
\left(\frac{a^{3}}{a^{3}+15 b c d}\right)^{\frac{1}{2}} \geqslant \frac{a^{x}}{a^{x}+b^{x}+c^{x}+d^{x}}
$$

the real number $x=$
5. $\frac{15}{8}$.

Rearranging the given inequality, we get
$$
\begin{array}{l}
\left(a^{x}+b^{x}+c^{x}+d^{x}\right)^{2}-a^{2 x} \geqslant 15 a^{2 x-3} b c d . \\
\text { The left side of equation (1) }=\left(b^{x}+c^{x}+d^{x}\right)\left(2 a^{x}+b^{x}+c^{x}+d^{x}\right) \\
\geqslant 15 \sqrt[3]{(b c d)^{x}} \cdot \sqrt[8]{a^{2 x}(b c d)^{x}} \\
=15 a^{\frac{2 x}{5}}(b c d)^{\frac{8 \pi}{13}} .
\end{array}
$$

Let $\frac{8 x}{15}=1$, we get $x=\frac{15}{8}$, at this point, $\frac{2 x}{5}=2 x-3=\frac{3}{4}$. Therefore, $x=\frac{15}{8}$.