6・167 $f$ 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
$$f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}}$$

Determine all such functions $f$.
［Solution］ Suppose $f$ satisfies the conditions given in the problem, then we have
$$f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}} .$$

Taking $u=\frac{1}{2}$, we get
$$f\left(x^{\frac{1}{2}} y^{v}\right) \leqslant f(x)^{\frac{1}{2}} f(y)^{\frac{1}{4 v}}$$

Choose $v$ such that
$$y^{v}=x^{\frac{1}{2}} \quad(x>1, y>1)$$

i.e.,
$$v=\frac{\ln x}{2 \ln y}$$

Thus,
$$\begin{array}{l}
f(x) \leqslant f(x)^{\frac{1}{2}} \cdot f(y)^{\frac{\ln y}{2 \ln x}} \\
f(x)^{\frac{1}{2}} \leqslant f(y)^{\frac{\ln y}{2 \ln x}} \\
f(x)^{\ln x} \leqslant f(y)^{\ln y}
\end{array}$$

By symmetry, we have $f(y)^{\ln y} \leqslant f(x)^{\ln x}$.
From (3) and (4), we get $f(x)^{\ln x}=f(y)^{\ln y}$.
This implies that $f(x)^{\ln x}$ is a constant $C(C>1)$, i.e.,
$$\begin{array}{l}
f(x)^{\ln x}=C \\
f(x)=C^{\frac{1}{\ln x}} \quad(C>1)
\end{array}$$

Conversely, if $f(x)=C^{\frac{1}{\ln x}}(C>1)$, then for $x, y>1, u, v>0$, we have
$$\begin{aligned}
f\left(x^{u} y^{v}\right) & =C^{\frac{1}{\ln \left(x^{u} y^{v}\right)}} \\
& =C^{\frac{1}{u \ln x+v \ln y}} .
\end{aligned}$$

Since $(u \ln x+v \ln y)^{2} \geqslant 4 u v \ln x \ln y$,
we have
$$\begin{aligned}
\frac{1}{u \ln x+v \ln y} & \leqslant \frac{u \ln x+v \ln y}{4 u v \ln x \ln y} \\
& =\frac{1}{4 v \ln y}+\frac{1}{4 u \ln x}
\end{aligned}$$

Thus,
$$\begin{aligned}
& f\left(x^{u} y^{v}\right) \\
= & C^{\frac{1}{u \ln x+v \ln y}} \\
\leqslant & C^{\frac{1}{4 u \ln x}}+\frac{1}{4 v \ln y} \\
= & C^{\frac{1}{4 u \ln x}} \cdot C^{\frac{1}{4 v \ln y}} \\
= & f(x)^{\frac{1}{4 u}} \cdot f(y)^{\frac{1}{4 v}}
\end{aligned}$$

Therefore, the function that satisfies the conditions given in the problem is
$$f(x)=C^{\frac{1}{\ln x}} \quad(C>0)$$