For example, $8 f$ is a function defined on $(1,+\infty)$ and takes values in $(1,+\infty)$, satisfying the condition: for any $x$, $y>1$, and $u, v>0$, we have $f\left(x^{4} \cdot y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} \cdot f(y)^{\frac{1}{4 v}}$. Determine all such functions $f$.
(CMO - 4 Problem)
Let $f$ satisfy the conditions given in the problem, then we have $f\left(x^{u} \cdot y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} \cdot f(y)^{\frac{1}{4 v}}$.
Taking $u=\frac{1}{2}$, we get $\quad f\left(x^{\frac{1}{2}} \cdot y^{v}\right) \leqslant f(x)^{\frac{1}{2}} \cdot f(y)^{\frac{1}{4 v}}$.
Choosing $v$ such that $y^{v}=x^{\frac{1}{2}}(x>1, y>1)$, i.e., $v=\frac{\ln x}{2 \ln y}$.
Thus, $f(x) \leqslant f(x)^{\frac{1}{2}} \cdot f(y)^{\frac{\ln y}{2 \ln x}}$,
which implies $f(x)^{\frac{1}{2}} \leqslant f(y)^{\frac{\ln y}{2 \ln x}}$, or equivalently, $f(x)^{\ln x} \leqslant f(y)^{\ln y}$.
By symmetry, we get $f(y)^{\ln y} \leqslant f(x)^{\ln x}$.
From (3) and (4), we have $f(x)^{\ln x} \leqslant f(y)^{\ln y}$.
This indicates that $f(x)^{\ln x}$ is a constant $c(c>0)$, i.e.,
$$
f(x)^{\ln x}=c, f(x)=c^{\frac{1}{\ln x}}(c>1) \text {. }
$$

Conversely, if $f(x)=c^{\frac{1}{\ln x}}(c>1)$, then for $x, y>1, u, v>0$, we have $f\left(x^{u} \cdot y^{v}\right)=c^{\frac{1}{\ln \left(x^{u} \cdot y^{v}\right)}}=c^{\frac{1}{u \ln x+v \ln y}}$.
Since, $(u \ln x+v \ln y)^{2} \geqslant 4 u v \cdot \ln x \cdot \ln y$,
it follows that $\frac{1}{u \ln x+v \ln y} \leqslant \frac{u \ln x+v \ln y}{4 u v \cdot \ln x \cdot \ln y}=\frac{1}{4 v \ln y}+\frac{1}{4 u \ln x}$.
Thus, $f\left(x^{u}, y^{v}\right)=c^{\frac{1}{u \ln x+v \ln y}} \leqslant c^{\frac{1}{4 u \ln x}+\frac{1}{4 v \ln y}}=c^{\frac{1}{4 u \ln x}} \cdot c^{\frac{1}{4 v \ln y}}=f(x)^{\frac{1}{4 u}} \cdot f(y)^{\frac{1}{4 v}}$.
Therefore, the function that satisfies the conditions given in the problem is $f(x)=c^{\frac{1}{\ln x}}(c>1)$.