Example 14 (4th China High School Mathematics Olympiad Winter Camp Problem) $f$ is a function defined on $(1,+\infty)$ and takes values in $(1,+\infty)$, satisfying the condition: for any $x, y>1$ and $u, v>0$, it holds that
$$
f\left(x^{u} y^{v}\right) \leqslant f^{\frac{1}{1 u}}(x) f^{\frac{1}{10}}(y),
$$

Determine all such functions $f$.
Solve: First, transform (1) into a univariate function. Let $x=y, u=v=\frac{t}{2}$, then for all $x>1, t>0$, we have $f\left(x^{t}\right) \leqslant f^{\frac{1}{1}}(x)$.
$x^{\frac{1}{t}} \rightarrow x$, we get $f(x) \leqslant f^{\frac{1}{t}}\left(x^{\frac{1}{t}}\right)$. Let $\frac{1}{t} \rightarrow t$, we get $f\left(x^{t}\right) \geqslant f^{\frac{1}{t}}(x)$, so for all $x>1, t>0$, we have $f\left(x^{t}\right)=f^{\frac{1}{2}}(x)$.
Take $x=\mathrm{e}, u=\mathrm{e}^{t}$, we get $f(u)=f^{\frac{1}{\ln u}}(\mathrm{e})$.
Let $a=f(\mathrm{e})>1, x \rightarrow u$, we get $f(x)=a^{\frac{1}{a x}}, a>1$.
Now verify that such a function indeed meets the requirements.
For any $x, y>1$ and $u, v>0$, we have
$f\left(x^{u} y^{v}\right)=a^{\frac{1}{\ln x^{u} y^{v}}}=a^{\frac{1}{u \ln r+v \ln y}}, f^{\frac{1}{4 u}}(x) f^{\frac{1}{4 v}}(y)=a^{\frac{1}{4 \mu \ln x}+\frac{1}{4 v \ln y}}$, we only need to prove
$$
\frac{1}{u \ln x+v \ln y} \leqslant \frac{1}{4 u \ln x}+\frac{1}{4 v \ln y} \Leftrightarrow(u \ln x)^{2}+2 u v \ln x \ln y+(v \ln y)^{2} \geqslant 4 u v \ln x \ln y
$$

which is $(u \ln x-v \ln y)^{2} \geqslant 0$, obviously true.