10. If for all positive real numbers $a, b, c, d$, the inequality $\left(\frac{a^{3}}{a^{3}+15 b c d}\right)^{\frac{1}{2}} \geqslant \frac{a^{x}}{a^{x}+b^{x}+c^{x}+d^{x}}$ always holds, find all real numbers $x$ that satisfy the condition.
10. Solution: Let $b=c=d$, and let $\frac{b}{a}=t$, then the original inequality transforms into $\left(\frac{a^{3}}{a^{3}+15 b^{3}}\right)^{\frac{1}{2}} \geqslant \frac{a^{x}}{a^{x}+3 b^{x}}$, which is $\left(\frac{1}{1+15 t^{3}}\right)^{\frac{1}{2}} \geqslant \frac{1}{1+3 t^{x}}$, which is $\left(1+3 t^{x}\right)^{2} \geqslant 1+15 t^{3}$, which is $2 t^{x-3}+3 t^{2 x-3} \geqslant 5$ for all positive $t$. Let $f(t)=2 t^{x-3}+3 t^{2 x-3}$. We will discuss it below.
If $x>3$, then $x-3>0, 2 x-3>0$, taking $t=1$, then $f(t)\left(\frac{2(3-x)}{3(2 x-3)}\right)^{\frac{1}{x}}$

when, $f^{\prime}(t)>0$; when $t<\left(\frac{2(3-x)}{3(2 x-3)}\right)^{\frac{1}{x}}$, $f^{\prime}(t)<0$.
Therefore, $t_{0}=\left(\frac{2(3-x)}{3(2 x-3)}\right)^{\frac{1}{x}}$ is the unique minimum value of $f(t)$ on $(0,+\infty)$.
When $x \neq \frac{15}{8}$, $t_{0} \neq 1$, so $f\left(t_{0}\right)<f(1)=5$, contradiction!
When $x=\frac{15}{8}$, $b c d \leqslant\left(\frac{b+c+d}{3}\right)^{3}$, by Jensen's inequality: $b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}} \geqslant 3\left(\frac{b+c+d}{3}\right)^{\frac{15}{8}}$, so, when $(b, c, d) \rightarrow\left(\frac{b+c+d}{3}, \frac{b+c+d}{3}, \frac{b+c+d}{3}\right)$, the left side of the $f(t)$ expression decreases, and the right side increases, hence it is sufficient to prove the case $b=c=d$. At this point, the $f(t)$ expression transforms into $\frac{2}{t^{\frac{9}{8}}}+3 t^{\frac{6}{8}} \geqslant 5$, and the left side $=\frac{1}{t^{\frac{9}{8}}}+\frac{1}{t^{\frac{2}{8}}}+t^{\frac{6}{8}}+t^{\frac{6}{8}}+t^{\frac{6}{8}} \geqslant 5=$ right, hence the only solution is $x=\frac{15}{8}$.