1. For all positive real numbers $a, b, c, d$ we have $\left(\frac{a^{3}}{a^{3}+15 b c d}\right)^{\frac{1}{2}} \geq \frac{a^{x}}{a^{x}+b^{x}+c^{x}+d^{x}}$, then $x=$
1. $\frac{15}{8}$

Analysis: By the AM-GM inequality $b^{x}+c^{x}+d^{x} \geq 3(b c d)^{\frac{x}{3}}$, it is sufficient to prove:
$$
\left(\frac{a^{3}}{a^{3}+15 b c d}\right)^{\frac{1}{2}} \geq \frac{a^{x}}{a^{x}+3(b c d)^{\frac{x}{3}}} \Leftrightarrow 6 a^{3+x}(b c d)^{\frac{x}{3}}+9 a^{3}(b c d)^{\frac{2 x}{3}} \geq 15 a^{2 x} b c d
$$
$6 a^{3-x}(b c d)^{\frac{x}{3}-1}+9 a^{3-2 x}(b c d)^{\frac{2 x}{3}-1} \geq 15$ . Using the method of undetermined coefficients, take $\lambda, \mu \in N_{+}$, by the AM-GM inequality:
$$
\begin{array}{l}
6 a^{3-x}(b c d)^{\frac{x}{3}-1}+9 a^{3-2 x}(b c d)^{\frac{2 x}{3}-1}=\lambda \cdot \frac{6}{\lambda} \cdot a^{3-x}(b c d)^{\frac{x}{3}-1}+\mu \cdot \frac{9}{\mu} \cdot a^{3-2 x}(b c d)^{\frac{2 x}{3}-1} \\
\geq(\lambda+\mu) \cdot\left(\left(\frac{6}{\lambda}\right)^{\lambda} \cdot\left(\frac{9}{\mu}\right)^{\mu} a^{3 \lambda+3 \mu-(\lambda+2 \mu) x}(b c d)^{\frac{(\lambda+2 \mu) x}{3}-\lambda-\mu}\right)^{\frac{1}{\lambda+\mu}}
\end{array}
$$

It is sufficient that $\lambda, \mu$ satisfy:
$$
\left\{\begin{array} { c } 
{ ( \frac { 6 } { \lambda } ) ^ { \frac { \lambda } { \lambda + \mu } } \cdot ( \frac { 9 } { \mu } ) ^ { \frac { \mu } { \lambda + \mu } } \cdot ( \lambda + \mu ) = 1 5 } \\
{ 3 \lambda + 3 \mu - ( \lambda + 2 \mu ) x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\lambda=2 \\
\mu=3 \\
x=\frac{15}{8}
\end{array}\right.\right.
$$

Thus, the original inequality holds, hence $x=\frac{15}{8}$